Question

A copper calorimeter of mass $1.5kg$ has $200g$ 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 $=390J/kg-{}^{\circ }C)$

• 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
$Q_{\text{lost }}=(mg)\left(1\frac{\text{ cal }}{g-{}^{\circ }C}\right)(50-40){}^{\circ }C$ ...(i)
Heat gained by the calorimeter
$Q_1=\left[(1500g)\left(\frac{390}{4.2\times 10^3}\frac{cal}{g-{}^{\circ }C}\right)\right](40-25{)}^{\circ }C$
Heat gained by the water present in it
$Q_2=\left[(200g)\left(1\frac{cal}{g-{}^{\circ }C}\right)\right](40-25{)}^{\circ }C$
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