Question

A copper calorimeter of mass $1.5\, kg$ has $200\, g$ of water at $25^{\circ} C$. How much water (approx.) at $50^{\circ} C$ is required to be poured in the calorimeter, so that the equilibrium temperature attained is $40^{\circ} C ?$ (Sp. heat capacity of copper $\left.=390\, J / kg -{ }^{\circ} C \right)$

• A. 508.3 gram
• B. 450 gram
• C. 370 gram
• D. 610 gram

Answer 1

Answer: (A)

Let ' $m$ ' $g$ be the required mass of water.
Then heat lost by hot water
$Q_{\text {lost }}=(m g)\left(1 \frac{\text { cal }}{ g -{ }^{\circ} C }\right)(50-40){ }^{\circ} C$ ...(i)
Heat gained by the calorimeter
$Q_{1}=\left[(1500 g )\left(\frac{390}{4.2 \times 10^{3}} \frac{ cal }{ g -{ }^{\circ} C }\right)\right](40-25)^{\circ} C$
Heat gained by the water present in it
$Q_{2}=\left[(200 g )\left(1 \frac{ cal }{ g -{ }^{\circ} C }\right)\right](40-25)^{\circ} C$
Total heat gained by the calorimeter and the water present in it
$Q_{\text {gain }}=Q_{1}+Q_{2}$ ...(ii)
From principle of calorimetry, heat lost by hot body = heat gained by cold body So, equating Eqs. (i) and (ii), and solving for $m$, we get
$m=508.93$ gram