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प्रश्न
A body cools down from 50°C to 45°C in 5 mintues and to 40°C in another 8 minutes. Find the temperature of the surrounding.
उत्तर
Let the temperature of the surroundings be T0°C.
Case 1:
Initial temperature of the body = 50°C
Final temperature of the body = 45°C
Average temperature = 47.5 °C
Difference in the temperatures of the body and its surrounding = (47.5 − T)°C
Rate of fall of temperature = `(Deltat)/t = (5)/5 = 1°`C/min
By Newton's law of cooling,
`(dt)/dt = -K [T_{avg} - T_0 ]`
1 = -K [47.5 - t0]........... (i)
Case 2:
Initial temperature of the body = 45°C
Final temperature of the body = 40°C
Average temperature = 42.5 °C
Difference in the temperatures of the body and its surrounding = (42.5 − T0)°C
Rate of fall of temperature = `(DeltaT)/t = 5/8 = 5/8`°C/min
From Newton,s law of cooling,
`(dT)/(dt) = -K[T_{avg} - T_0 ]`
0.625 = -K [42.5 - T0] ........ (2)
Dividing (1) by (2),
`1/0.625 = (47.5 - T_0)/(42.5 - T_0)`
⇒ 42.5 - T_0 = 29.68 - 0.625T
⇒ T0 = 34° C
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