Test Papers – ICSE – Class – X
JEE MAIN ADVANCE 11th PCM Physics Heat & Thermodynamics
Hello, students, in this module we are going to look at a few important terms related to Calorimetry and the very important principle of Calorimetry based on which we have a number of questions coming up in the further modules. So, I want you to pay very, very close attention to this particular module because the concepts that we learn out here is going to be directly applied in the numericals.
So, to begin with, the important terms with Calorimetry. The first term that I will be encountering is Thermal Capacity. Now, what exactly is the Thermal Capacity? Now Thermal Capacity is defined as amount of heat required in order to raise the temperature of the entire body by 1 degree centigrade. Whereas now, if I have to compare this with specific heat capacity, it is the amount of heat required to raise the temperature of unit mass of a substance by 1 degree. So, the difference between thermal capacity and specific heat capacity comes in the fact that it is for the entire body and this is for a unit mass. So, if I have the relation here which will give me delta Q is equal to mC delta T, hence it can also be written as what, H into delta T. So, if I cancel delta T from both sides of the equation, I get my thermal capacity as the product of mass into specific heat capacity. So, this is a very important relation to be remembered, my dear students. The unit of thermal capacity is nothing but Joules per degree centigrade or Joules per Kelvin.
Now, in this list the next important term that we have is the water equivalent. Dear students, let me tell you water equivalent generally appears in the numerical and I don’t want anyone of you to get confused because when we have a mixture present in a calorimeter, they don’t mention to us about the mass of the calorimeter or its specific heat capacity. They give us directly what is the water equivalent of the calorimeter. So, from there you should not be confused, you should be able to use the concept of water equivalent very nicely to find out what will be the raise in the temperature of the calorimeter. So, here we try to understand, what exactly is meant by calorimeter? Let’s say, we have a block whose mass is m and its specific heat capacity is C and the change in temperature is delta T. So, how much heat will be related to this particular block. Delta Q is equal to mC delta T. Now let’s say the same amount of heat I use to raise the temperature of water by an equal difference. So, the same amount of heat for raising the temperature of water by equal level then how much water should I be taking that is known as the water equivalent. So, I can write down this delta Q is nothing but equals to W into 1 into delta T where the delta T will get cancelled from both sides of the equation. Hence, I will get the value of my value equivalent is equal to nothing but product of mass into specific heat capacity.
Having done this, let’s just go ahead with this to look at the basic principles of Calorimetry. What does it say? Let’s say we have a block which has got a mass m1, specific heat capacity C1. It is at a temperature T1 degree centigrade. We have got another block, whose mass is m2, its specific heat capacity is C2 and it is ata temperature T2. Now, given is temperature T1 is greater than T2. So, heat is going to flow from the body at a higher temperature at the body at a lower temperature and this is the direction of the heat flow. Now, after some time there is going to be a state of thermal equilibrium where the temperature of both the bodies is going to be equal. So the principle of Calorimetry states that heat lost by a body at higher temperature is equal to the heat gained by the body at a lower temperature. So, having done that, what is the amount of heat lost by the body at a higher temperature? It is m1C1 into T1 minus T that is the temperature difference for the body at a higher temperature. And what is the heat gained by a body at the lower temperature? It is m2C2 into T minus T2. Now, equating them this is what is the most important principle that we are learning in this particular module. Now, having done that, T is the temperature of equilibrium and if we solve it, we get the temperature of equilibrium as m1C1T1 plus m2C2T2 divided by m1C1 plus m2C2.
So having done these, students, we will be using the same concept in the numericals ahead.
Thank you very much.
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JEE MAIN ADVANCE 12th PCM Chemistry Demo Videos
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JEE MAIN ADVANCE 12th PCM Physics Alternating Current 2 Demo Videos
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JEE MAIN ADVANCE 12th PCM Maths Definite Integration Demo Videos
Hello, students, let us take some examples on properties. So the first example says integration from minus 1 to 1 bracketed of x bracket of 1 plus sin pi x plus 1 dx.
Now over here in the example, students, we are going to use this property that is property number 3, that is I am going to break this integral into two parts. So let us break this integral. That is first integral is from minus 1 is equal to 0 f(x)dx plus integral from 0 to 1 f(x) dx. Now, students, note over here that in the first integral limits are given from minus 1 to 0. So 1 x is between minus 1 to 0. I am going to calculate sin pi x, so sin pi x will be from minus 1 to 0 because pi x, when x is between minus 1 to 0, pi x will be in third and fourth quadrant. And in third and fourth quadrant, sin pi x is from minus 1 to 0. And hence when I add 1 both the sides it will become 1 plus sin x will be from 0 to 1, substituting this using the bracketed value I add 1 plus sin pi x will be equal to zero. Because 1 plus sin pi x was from 0 to 1, so bracketed value was becoming equal to zero. Now substituting this value in the integral you can see over there the value becomes zero of the bracketed value. And now for the second integral limits are given from 0 to 1, so 1x between 0 to 1, sin pi x, since pi x is from 0 to pi, sin pi x will become 0 to 1. And now when I take, add 1 both the the side I will get 1 to 2, the range will become from 1 to 2. So bracketed of this value will become equal to 1. Now I am going to substitute this value 1 over there in the integral so it will become 1 over there in the integral. Now simplifying it further I will obtain I will be equal to integral minus 1 equal to 0. Now since the inner bracket was become equal to 0 so the integral of bracketed 1, bracket 1 is simple 1, so it becomes 1 dx. And for the second integral, it’s inner bracket has become equal to 1 but x is still over there so it becomes bracket of x plus 1.
Now, students, when I simplify it further, let us check how do we do it. For the first integral the value is very easy to simplify, that is integrating 1, we get x limits minus 1 to 0. For second integral, students, I hope you remember this that bracketed of x plus integer will be equal to bracketed x plus integer. Over here in the second integral bracket x plus 1 is there, 1 is the integer. So when I simplify it becomes 0 to 1 bracketed plus 1 dx. In the integral of bracketed limits are given from 0 to 1. Now since limit is from 0 to 1, bracketed x will become equal to 0. Hence the second integral becomes 0. And now when I simplify it further I get the answer as 2.
So, students, I hope you understood the question. Let’s move further.
Now the second integral, the second question which is given to us is I have to evaluate the value of this integral. You can see it is given in two parts. In both the parts limits are different. In first integral, limit is from minus 4 to minus 5, in second integral limit is from minus 2 to 1. Looking at this I can say we will have to use the substitution which we have studied, x is equal to b minus a into t plus a, that is this property which we studied. I convert both the limits from 0 to 1. So to do that let’s start with the first integral I have taken it as I1. So minus 4 to minus 5 sin of x square minus 3. Now substituting the limit, substituting x equal to b minus a into t plus a, over here b minus a that is upper limit minus lower limit plus a is the lower limit that is minus 4. So simplifying it x becomes minus t minus 4. And now when I substitute it in the given integral I get integral 0 to 1 using the property, students, x is replaced by minus t minus 4, taking the square of it, simplifying it further. When I simplify it further I expand the bracket, right, and then I am simplifying it and I1 becomes 0 to 1, sin of t square plus 8t plus 13. I have taken this integral aside. And I am now going to solve integral second that is I2, which is given as minus 2 to 1 integration of this. Again substituting x equal to b minus a into t plus a. b and a are the upper and the lower limits. So when I substitute the upper and lower limits over here x equals to t minus 2. Substituting this substitution in the integral I get 0 to 1 sin of substitution. And now simplifying it further I get I2 equal to 0 to 1 sin of the integral. And now simplifying it further 0 to 1 it is just a simplification I have done and obtained this.
You can see over here, friends, I1 and I2 are both of opposite signs. So when I add I1 and I2 it becomes equal to 0. So, students, the value of this I1 plus I2 becomes equal to 0.
I hope you understood this, thank you.
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State Board English,Std VIII Science Demo Video
Hello, students, today we will start with new a chapter, Agriculture.
Many plants grow because their seeds are widely dispersed in nature. These seeds are carried away by air. In the picture, you can see dandelion seeds carried away by air.
Next, is birds. In the picture, you can see are dropped by birds.
Next is water. In the picture, you can see coconut seed carried away by water to the shore.
Next is by insects. Pollen grains of the flowers stick to the insect’s body. These pollen grains fall off wherever the insect is travelling.
Seeds of trees like banyan and peepul are dispersed through the droppings of the birds. In the picture, you can see birds eating fruits of the banyan and peepul tree which are dispersed by their dropping that is through their waste matter.
So, the seeds are carried away by air, birds, water and insects.
Now, let’s see Nature Nurturing the Plants. How does nature nurture the plants? Seeds take roots easily if the soil is fertile. How does the soil become fertile? Remains of dead animals, dry leaves, droppings of animals get buried in the soil and are transformed into manure. So, dead animals, dry leaves, horse manure makes the soil fertile. Then rain water also helps the growth of the seed. After that under favourable conditions, these seeds will take roots and grow into trees. Thus, plants are nurtured by nature itself.
Now, let’s see the past versus present scenario of the forests. Many, many years ago there were thick forests on the earth. But now, man cleared forests to perform agriculture. That is to grow variety of crops to meet his food requirements and to make settlements.
Agriculture for food crops. Agricultural produce includes cereal foods. Cereal foods include rice, jowar, maize, wheat, bajra and ragi.
It also includes some pulses like chana also called as Bengal gram, udid called as black gram, masoor called as lentils, moong called as green gram, tur called as arhar and matki also called as moth beans.
It also includes oilseeds. Now, what are oilseeds? Seeds from which oil is extracted are oil seeds. It includes oilseeds like sunflower, groundnut, linseed or javas, safflower, sesame or til and soya bean.
It also includes some vegetables like cluster beans called as gavar, tomatoes, colocasia or arwi, bitter gourd or karela, brinjal, chakwat, snake gourd also called as padval and fenugreek called as methi.
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State Board English-Std VIII Mathematics Demo video
Hello, students, today we are going begin new chapter that is Cube and cube roots, a very lovely chapter. Let’s proceed ahead students and study about cube and cube roots.
So, students we have to now find out the cube of 2. In order to find cube of a number just multiply the number 3 times. So the number over here is 2 and to find the cube of 2 just multiply the number 3 times. So, what are we going to do over here, we will multiply the number 3 times, 2 multiplied by 2 multiplied by 2, which is equal to 8. Now, students, this can also be written in index form as 2 raise to 3. So, students 2 raise to 3 is equal to 8.
Proceeding ahead students, how to find the cube of a number? We just found the cube of the number. But we will now go step by step. So, find the cube of 3 that is 3 raise to 3. So, first step will be to find the cube of 3, multiply 3 three times. So, what will happen over here, 3 multiplied by 3 multiplied by 3. Now, students 3 multiplied 3 will be equal to 9, 9 will be again multiplied by 3. So, 9 multiplied by 3 will be equal to 27. Students, cube of 3 is written as 3 raise to 3 is equal to 27 or this means cube of 3 is equal to 27.
Moving ahead, students, with exercise number 45. The first question says, evaluate the following and write which number is the cube of which other number. So, the first number given to us is 3 and we have to find out the cube of 3. So, solution, 3 raise to 3, we will have to multiply 3 three times. So, students this will be written as 3 into 3 into 3, 3 multiplied by 3 is equal to 9, 9 multiplied 3 is equal to 27. So, students, 27 is the cube of 3.
Moving ahead with the next question, where we have to find the cube of 7. So, this is written as 7 raise to 3. We will have to multiple 7 three times. So, students this will be written as 7 multiplied 7 multiplied by 7. 7 multiplied 7 is equal to 49, 49 multiplied by 7 is equal to 343. So, students, 343 is the cube of 7.
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