Solution:

Let the temperature of the surroundings be \( T_s \).

Given:
- First cooling: from \( 62^\circ \text{C} \) to \( 50^\circ \text{C} \) in 10 minutes.
- Second cooling: from \( 50^\circ \text{C} \) to \( 42^\circ \text{C} \) in the next 10 minutes.

Step 1: Use Newton's Law of Cooling
According to Newton's law:
\[
\frac{T_1 - T_2}{T_1 - T_s} = \frac{T_2 - T_3}{T_2 - T_s}
\]

Step 2: Substitute Values
Let \( T_1 = 62^\circ \text{C} \), \( T_2 = 50^\circ \text{C} \), and \( T_3 = 42^\circ \text{C} \).

Then:
\[
\frac{62 - 50}{62 - T_s} = \frac{50 - 42}{50 - T_s}
\]

Step 3: Solve for \( T_s \)
\[
\frac{12}{62 - T_s} = \frac{8}{50 - T_s}
\]

Cross-multiplying:
\[
12(50 - T_s) = 8(62 - T_s)
\]

Expanding and simplifying:
\[
600 - 12T_s = 496 - 8T_s
\]

\[
4T_s = 104 \Rightarrow T_s = 26^\circ \text{C}
\]

So, the temperature of the surroundings is \( 26^\circ \text{C} \).