Question

A hot body at temperature ${\theta _1}$ is mixed with a cold body at temperature ${\theta _2}$. The fall in temperature of the hot body is equal to the rise in temperature of the cold body. The material and the mass of both the bodies are the same. Then find the resultant temperature.

Answer 1

Hint: Here, we will proceed by writing down the amount of the heat removed from the hot body and the amount of the heat absorbed by the cold body. Then, we will equate both of these to find the value of the resultant temperature.

Formula Used: H = mc$\Delta {\text{T}}$.

Complete Step By Step Answer:

Given, Initial temperature of hot body = ${\theta _1}$

Initial temperature of cold body = ${\theta _2}$

Fall in the temperature of the hot body = Rise in the temperature of the cold body

The above equation represents that both the hot and cold bodies are coming in equilibrium with each other and let us assume that the final resultant temperature of both the bodies be T

Mass of the hot body = Mass of the cold = m (assume)

It is also given that the material of both the hot and cold bodies are the same. Due to this the specific heat of both the hot and cold bodies will be the same.

Let us assume c be the specific heat of these bodies.

i.e., Specific heat of hot body = Specific heat of cold body = c

As we know that the heat transfer is given by

H = mc$\Delta {\text{T }} \to {\text{(1)}}$ where H represents the amount of heat transfer, m denotes the mass of the body, c denotes the specific heat of the body and $\Delta {\text{T}}$ denotes the change in temperature

Using the formula given by equation (1) for the hot body, we can write

Heat removed from the hot body H = mc$\left( {{\theta _1} - {\text{T}}} \right)$

Using the formula given by equation (1) for the hot body, we can write

Heat absorbed by the cold body H = mc$\left( {{\text{T}} - {\theta _2}} \right)$

Since, the heat removed from the hot body will be equal to the heat absorbed by the cold body

i.e., Heat removed from the hot body = Heat absorbed by the cold body

$ \Rightarrow $ mc$\left( {{\theta _1} - {\text{T}}} \right)$ = mc$\left( {{\text{T}} - {\theta _2}} \right)$

$   \Rightarrow {\theta _1} - {\text{T}} = {\text{T}} - {\theta _2} $

$   \Rightarrow {\text{T}} + {\text{T}} = {\theta _1} + {\theta _2} $

$   \Rightarrow 2{\text{T}} = {\theta _1} +{\theta _2} $

$  \Rightarrow {\text{T}} = \dfrac{{{\theta _1} + {\theta _2}}}{2}  $

Therefore, the resultant temperature is $\dfrac{{{\theta _1} + {\theta _2}}}{2}$.

Note: It is important to note that whenever a hotter body comes in contact with a colder body, the direction of heat flow is from hotter body to colder body resulting in the decrease and increase of the temperatures of the hot and cold bodies respectively. Here, complete equilibrium will be established when the final temperature of both the bodies will be equal.