Question

A copper calorimeter contains $50g$ of water at $16°C$, when $40g$ of water at $36°C$ is added, the resulting temperature of the mixture is $24°C$. Calculate the heat capacity of the calorimeter.

Answer 1

Step 1: Organised table of the given data

Substance	Mass	S.H.C	Initial Temperature	Final Temperature=$24°C$
Hot water	$40g$	$4.2J{g}^{-1}°C^{-1}$	$36°C$	${\theta }_F=36-24=12°C$
Cold water	$50g$	$4.2J{g}^{-1}°C^{-1}$	$16°C$	${\theta }_R=24-20=4°C$
Vessel	?	?	$4°C$

Step 2: Calculating the heat energy

Let $mc=x$ be the heat capacity of the calorimeter

resulting temp. of mixture $24°C$

on adding hot water at $36°C$ when added to cold water at $16°C$, hot water loses heat and cold water and the calorimeter gains heat till the temp. becomes $24°C$.

$\text{Heat gained by calorimeter}+\text{cold water=}=(x+50\times 4.2)(24-16)\mathrm{J}$

$$\begin{gathered} \text{Heat lost by hot water =mc(dT)} \\ =(40\times 4.2)\text{fall in temp.} \\ \begin{array}{r}=(4\times 42)(36-24)\\ =168\times 12\mathrm{J}\end{array} \end{gathered}$$

$$\begin{gathered} \text{Heat gained}=\text{Heat lost} \\ (x+210)8=168\times 12 \\ x+210=\frac{168\times 12}{8} \\ x+210==21\times 12 \\ x=252-210 \\ x=42{\mathrm{J}}^{\circ }{\mathrm{C}}^{-1} \end{gathered}$$

Hence, the thermal of the calorimeter is $42{\mathrm{J}}^{\circ }{\mathrm{C}}^{-1}.$