A bodyat 50 degreescoolsin surroundingmaintainedat30 degreesthe temperatureat whichthe rate ofcooling ishalf thatof thebeginning

We will solve this problem using Newton’s Law of Cooling, which states:

Rate of cooling = k (T - T_s)

where

T is the temperature of the body at any time t,
T_s is the surrounding temperature,
k is a constant.
Step 1: Define Initial Conditions
Initial temperature of the body, T_0 = 50°C
Surrounding temperature, T_s = 30°C
Initial rate of cooling:
(dT/dt)₀ = k (50 - 30) = 20k
Step 2: Find Temperature When Rate of Cooling is Half
We need to find the temperature T at which the rate of cooling is half of the initial rate.

So,
(dT/dt) = (1/2) * (dT/dt)₀
k (T - 30) = (1/2) * (20k)
T - 30 = (1/2) * 20
T - 30 = 10
T = 40°C

Final Answer:
The temperature at which the rate of cooling is half of the initial rate is 40°C.