A mass of 50 g of water in a closed vessel, with surroundings at a constant temperature from 30°C to 25°C takes 2 minutes to cool. A mass of 100 g of another liquid in an identical vessel with identical surroundings takes the same time to cool from 30°C to 25°C. The specific heat of the liquid is : (The water equivalent of the vessel is 30 g.)

We are given two cooling experiments: one with water and one with another liquid. Both are in identical vessels with identical surroundings and cool from 30°C to 25°C in the same time of 2 minutes. The water equivalent of the vessel is 30 g, meaning the vessel behaves as if it were 30 g of water in terms of heat capacity.

First, recall Newton's law of cooling: the rate of cooling is proportional to the temperature difference between the body and its surroundings. For both systems, the initial and final temperatures are the same, and the cooling time is identical. This implies that the cooling curves are identical, leading to the same decay constant in the exponential cooling equation.

The decay constant $$b$$ is given by $$b = \frac{k}{C}$$, where $$k$$ is a constant depending on the surface area and conditions (same for both vessels), and $$C$$ is the total heat capacity of the system (including the vessel). Since the cooling curves are identical, $$b$$ must be the same for both systems. Therefore, $$\frac{k}{C_{\text{water}}} = \frac{k}{C_{\text{liquid}}}$$, which implies $$C_{\text{water}} = C_{\text{liquid}}$$.

Now, calculate the total heat capacity for the water system. The specific heat of water is $$1 \, \text{cal/g°C}$$. The mass of water is 50 g, and the water equivalent of the vessel is 30 g. The heat capacity of the water is $$50 \times 1 = 50 \, \text{cal/°C}$$, and the heat capacity of the vessel is $$30 \times 1 = 30 \, \text{cal/°C}$$. Thus, the total heat capacity for the water system is:

$$C_{\text{water}} = 50 + 30 = 80 \, \text{cal/°C}$$

For the liquid system, let the specific heat of the liquid be $$c_l \, \text{cal/g°C}$$. The mass of the liquid is 100 g, so the heat capacity of the liquid is $$100 \times c_l \, \text{cal/°C}$$. The vessel has the same water equivalent of 30 g, contributing $$30 \times 1 = 30 \, \text{cal/°C}$$. Thus, the total heat capacity for the liquid system is:

$$C_{\text{liquid}} = 100c_l + 30 \, \text{cal/°C}$$

Since $$C_{\text{water}} = C_{\text{liquid}}$$, we set them equal:

$$80 = 100c_l + 30$$

Solve for $$c_l$$:

Subtract 30 from both sides:

$$80 - 30 = 100c_l$$

$$50 = 100c_l$$

Divide both sides by 100:

$$c_l = \frac{50}{100} = 0.5 \, \text{cal/g°C}$$

The options are in kcal/kg. Note that $$1 \, \text{cal/g°C} = 1 \, \text{kcal/kg°C}$$ because:

• 1 kcal = 1000 cal
• 1 kg = 1000 g

Thus, $$0.5 \, \text{cal/g°C} = 0.5 \, \text{kcal/kg°C}$$.

Comparing with the options:

A. 2.0 kcal/kg

B. 7 kcal/kg

C. 3 kcal/kg

D. 0.5 kcal/kg

Hence, the correct answer is Option D.