Question

When $0.4kg$ of water at ${30}^{\circ }C$ is mixed with $0.15kg$ of water at ${25}^{\circ }C$ contained in a calorimeter, the final temperature is found to be ${27}^{\circ }C$, find the water equivalent of the calorimeter.

• A. $1kg$
• B. $0.5kg$
• C. $1.5kg$
• D. $0.450kg$

Answer 1

The correct option is D $0.450kg$

Formula used: $Q=mL,Q=ms\Delta T$

Given: $m_1=0.15kg,T_1={25}^{\circ }C,m_2=0.4kg,T_2={30}^{\circ }C,T={27}^{\circ }C$

Let $s$ and $W$ be the heat capacity of water and the water equivalent of the calorimeter.
Heat gained by (water + calorimeter) at temperature $T_1=$ Heat lost by water at temperature $T_2$.

$m_{1}s(T-T_1)+Ws(T-T_1)=m_{2}s(T_2-T)$
$W=m_2\left(\frac{T_2-T}{T-T_1}\right)-m_1$
$W=0.4\left(\frac{30-27}{27-25}\right)-0.15$
$W=0.45kg$

FINAL ANSWER: (a).