Question

A calorimeter contains 10 g of water at $20^{\circ}C$. The temperature falls to $15^{\circ}C$ in 10 min. When calorimeter contains 20 g of water at $20^{\circ}C$, it takes 15 min. for the temperature to become $15^{\circ}C$. The water equivalent of the calorimeter is}

• A. 50 g
• B. 25 g
• C. 10 g
• D. 5 g

Hint:

Rate of heat loss depends on the surface area and temperature difference, not the mass of liquid.

Answer 1

The Correct Option is C

Step 1: Formula
By Newton's law of cooling, $\frac{dQ}{dt} = K(\theta - \theta_0)$.
The heat lost is $Q = (ms + W)\Delta \theta$, where $W$ is the water equivalent.

Step 2: Analysis
Since the temperature range and surroundings are the same, the rate of heat loss $\frac{dQ}{dt}$ is constant.
- Case 1: $\frac{(10 + W) \times 1 \times 5}{10} = R$ - Case 2: $\frac{(20 + W) \times 1 \times 5}{15} = R$

Step 3: Calculation
$\frac{10+W}{10} = \frac{20+W}{15} \implies 3(10+W) = 2(20+W)$
$30 + 3W = 40 + 2W \implies W = 10$ g.

Step 4: Conclusion
Hence, the water equivalent is 10 g. Final Answer: (C)