Hello, students, in this module we are going to look at a few important terms related to calorimetry and a 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 of calorimetry the first term I will be encountering is thermal capacity. Now what exactly is thermal capacity, thermal capacity is defined as the amount of heat required in order to raise the temperature of the entire body by 1 degree centigrade. Whereas if I have now 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 this can also be written as what, H into delta T. So if I cancel the 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 we have is the water equivalent. Dear students, let me tell you water equivalent generally appears in the numerical and I don’t want any one 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 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 change in temperature is delta T. So how much heat will be related to this particular block. Delta Q is equal to mc plus Delta C. 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 water equivalent is equal to nothing but the product mass into specific heat capacity. Having done this let’s just go ahead to look at the basic principle of calorimetry. What does it say?

Let’s say we have a block which has got a mass m1, specific capacity is C1, it is at a temperature T1 degree centigrade. We have another block whose mass m2, its specific heat capacity is C2, and it is at a temperature T2. Now given is temperature T1 is greater than T2, so heat is going to flow from the body at a higher temperature to 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 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 a lower temperature, it is m2C2 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 this, students, we will be using this same concept in the numericals ahead.

Thank you very much.

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