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A body cools in 7 min from 70° to 50°C. What time does it take to cool from 50°C to 38°C, if the surrounding temperature is 20°C?
1. 3.5 min
2. 11 min
3. 7 min
4. 10 min

1 Answer

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Best answer
Correct Answer - Option 3 : 7 min

CONCEPT:

Newton’s law of cooling

  • According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body, and its surroundings.
  • Mathematically it is given as,

\(⇒ \frac{dT}{dt}=k(T_{t}-T_{s})\)

Where 

  • Tt = temperature at time t
  • Ts = temperature of the surrounding
  • k = Positive constant that depends on the area and nature of the surface of the body under consideration
  • dt = time required

CALCULATION:

For case 1:

⇒ Ts = 20°C

⇒ dT = 70 - 50

⇒ dT = 20°C

⇒ dt = 7 min

\(⇒ T_{t}=\frac{70+50}{2}\)

⇒ Tt = 60°C

\(⇒ \frac{dT}{dt}=k(T_{t}-T_{s})\)

\(⇒ \frac{20}{7}=k(60-20)\)

\(⇒ \frac{20}{7}=40k\)     -----(1)

For case 1:

⇒ Ts = 20°C

⇒ dT = 50 - 38

⇒ dT = 12°C

\(⇒ T_{t}=\frac{50+38}{2}\)

⇒ Tt = 44°C

\(⇒ \frac{dT}{dt}=k(T_{t}-T_{s})\)

\(⇒ \frac{12}{dt}=k(44-20)\)

\(⇒ \frac{12}{dt}=24k\)     -----(2)

By equation 1 and equation 2,

⇒ dt = 7 min

  • Hence, option 3 is correct.

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