Hot water cools from 60°C to 50°C in the first 10 minutes and to 42°C in the next 10 minutes. The temperature of the surroundings is:

According to Newton's law of cooling, the rate of cooling is proportional to the difference between the temperature of the body and the surroundings. The formula for the average rate of cooling over a time interval is given by:

$$\frac{T_1 - T_2}{t} = k \left( \frac{T_1 + T_2}{2} - T_s \right)$$

where $$ T_1 $$ is the initial temperature, $$ T_2 $$ is the final temperature, $$ t $$ is the time interval, $$ T_s $$ is the temperature of the surroundings, and $$ k $$ is a constant.

For the first 10 minutes:

• Initial temperature $$ T_1 = 60^\circ \text{C} $$
• Final temperature $$ T_2 = 50^\circ \text{C} $$
• Time $$ t = 10 $$ minutes
• Rate of cooling $$ \frac{60 - 50}{10} = 1^\circ \text{C per minute} $$
• Average temperature $$ \frac{60 + 50}{2} = 55^\circ \text{C} $$

So, the equation becomes:

$$1 = k (55 - T_s) \quad \text{(Equation 1)}$$

For the next 10 minutes:

• Initial temperature $$ T_1 = 50^\circ \text{C} $$
• Final temperature $$ T_2 = 42^\circ \text{C} $$
• Time $$ t = 10 $$ minutes
• Rate of cooling $$ \frac{50 - 42}{10} = 0.8^\circ \text{C per minute} $$
• Average temperature $$ \frac{50 + 42}{2} = 46^\circ \text{C} $$

So, the equation becomes:

$$0.8 = k (46 - T_s) \quad \text{(Equation 2)}$$

We now have two equations:

$$1 = k (55 - T_s)$$

$$0.8 = k (46 - T_s)$$

To solve for $$ T_s $$, divide Equation 1 by Equation 2:

$$\frac{1}{0.8} = \frac{k (55 - T_s)}{k (46 - T_s)}$$

Simplify the left side:

$$\frac{1}{0.8} = 1.25$$

So:

$$1.25 = \frac{55 - T_s}{46 - T_s}$$

Cross-multiply:

$$1.25 \times (46 - T_s) = 55 - T_s$$

Calculate $$ 1.25 \times 46 $$:

$$1.25 \times 46 = 1.25 \times (40 + 6) = (1.25 \times 40) + (1.25 \times 6) = 50 + 7.5 = 57.5$$

So:

$$57.5 - 1.25 T_s = 55 - T_s$$

Bring all terms involving $$ T_s $$ to one side and constants to the other. Add $$ 1.25 T_s $$ to both sides:

$$57.5 = 55 - T_s + 1.25 T_s$$

$$57.5 = 55 + 0.25 T_s$$

Subtract 55 from both sides:

$$57.5 - 55 = 0.25 T_s$$

$$2.5 = 0.25 T_s$$

Solve for $$ T_s $$:

$$T_s = \frac{2.5}{0.25} = \frac{2.5 \times 100}{0.25 \times 100} = \frac{250}{25} = 10$$

So, the temperature of the surroundings is $$ 10^\circ \text{C} $$.

Verification: Substitute $$ T_s = 10 $$ into Equation 1:

$$1 = k (55 - 10) = k \times 45 \implies k = \frac{1}{45}$$

Now substitute into Equation 2:

$$0.8 = \frac{1}{45} \times (46 - 10) = \frac{1}{45} \times 36 = \frac{36}{45} = \frac{4}{5} = 0.8$$

This matches, confirming the solution.

Hence, the correct answer is Option B.