Question

A solid of mass 150 g at 200 °C is placed in 0.4 kg of water at 20 °C till a constant temperature is attained. If the S.H.C of the solid is 0.5 Jg^-1 K^-1, find the resulting temperature of the mixture.

Answer 1

Given:

The mass of the solid, $M=150g$

The mass of the water, $m=0.4kg=400g$

The initial temperature of the solid $=200°C$

The initial temperature of the water $=20°C$

The specific heat capacity of the solid, $C=0.5J{g}^{-1}{K}^{-1}$

Finding the resultant temperature of the mixture:

Let the final temperature of the mixture be $x$

The specific heat capacity of water, $c=4.182J{g}^{-1}{K}^{-1}$

From the principle of calorimetry, we know that the heat lost by the hot body = the heat gained by the cold body.

So, the heat lost by the solid = the heat gained by the water

So, $MC(200-x)=mc(x-20)$

By substituting the corresponding values in the above expression, we get

$$\begin{gathered} 150\times 0.5\times (200-x)=400\times 4.182\times (x-20) \\ \Rightarrow 75\times (200-x)=1672.8(x-20) \\ \Rightarrow 15000-75x=1672.8x-33456 \\ \Rightarrow 15000+33456=1747.8x \\ \Rightarrow 48456=1747.8x \\ \Rightarrow x=27.7 \end{gathered}$$

Therefore, the final temperature of the mixture = $27.7°C$