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Question: A solid maintained at \({{t}_{1}}^{\circ }C\) is kept in an evacuated chamber at temperature \({{t}_...

A solid maintained at t1C{{t}_{1}}^{\circ }C is kept in an evacuated chamber at temperature t2C{{t}_{2}}^{\circ }C (t2C{{t}_{2}}^{\circ }C​<<​t1C{{t}_{1}}^{\circ }C). The rate of heat absorbed by the body is proportional to:

A. t24t14 B. (t24+273)(t14+273) C. t2t1 D. t22t12 \begin{aligned} & \text{A}\text{. }{t}_{2}^{4}-{t}_{1}^{4} \\\ & \text{B}\text{. }\left( {t}_{2}^{4}+273 \right)-\left( {t}_{1}^{4}+273 \right) \\\ & \text{C}\text{. }{{t}_{2}}-{{t}_{1}} \\\ & \text{D}\text{. }{t}_{2}^{2}-{t}_{1}^{2} \\\ \end{aligned}

Explanation

Solution

First write the values that are given. Then write the formula for rate of cooling. Figure out the constant values in it. By figuring out the constant values we can identify on which components of the body the rate of heat absorbed is dependent. When we figure it out the components we have to see which of the components are given in the question. If the components are available in the question then replace them in the equation to find the answer.

Formula Used:
dtdT=bA(tt0)\dfrac{dt}{dT}=bA(t-{{t}_{0}})

Complete step by step answer:
This question has been taken from the chapter Newton's laws of cooling.
Here t1C{{t}_{1}}^{\circ }C is the initial temperature of the body.
And t2C{{t}_{2}}^{\circ }C is the final temperature of the body.
Further it is said that, (t2C{{t}_{2}}^{\circ }C​<<​t1C{{t}_{1}}^{\circ }C).
We know that the rate of cooling is,
dtdT=bA(tt0)\dfrac{dt}{dT}=bA(t-{{t}_{0}}),
So, we can say that the rate of cooling is directly proportional to,
dtdT(tt0)\dfrac{dt}{dT}\propto \left( t-{{t}_{0}} \right) ,
Now , replacing the values of the equation above with the values given in the question.
dtdT(t2t1)\dfrac{dt}{dT}\propto \left( {{t}_{2}}-{{t}_{1}} \right).
So we can say that the rate of cooling or the rate of heat absorbed by the body which is cooling is proportional to (t2t1)\left( {{t}_{2}}-{{t}_{1}} \right).
So, the correct option is option C.

Note:
In the equation dtdT=bA(tt0)\dfrac{dt}{dT}=bA(t-{{t}_{0}}), ‘bA’ is a constant, ‘t’ is the final temperature and t0{{t}_{0}} is the initial temperature. So the constant part is getting omitted because it is always the same and the value will not depend on it and as a result the whole rate of cooling gets dependent on the final and the initial temperature.