The temperature of the body drops from 60°C to 40°C in 7 min. The surrounding temperature is 10°C. The temperature of the body drops from 40°C to T°C in 7 min. Find the value of T
The temperature of the body drops from 60°C to 40°C in 7 min. The surrounding temperature is 10°C. The temperature of the body drops from 40°C to T°C in 7 min. Find the value of T
16
20
28
30
The Correct Option is C
Solution and Explanation
The correct answer is option (C): 28
\(\frac{60-40}{7}=K(50-10)\)
\(\frac{(40-T)}{7}=K(\frac{40+T}{2}-10)\)
\(\Rightarrow \frac{20}{40-T}=(\frac{40\times 2}{T+20})\)
\(\Rightarrow T+20=160-4T\)
\(\Rightarrow 5T=140\)
\(T=\frac{140}{5}=28^{\circ}\)C
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Concepts Used:
Newton’s Law of Cooling
Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings.
Derivation of Newton’s Law of Cooling
Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings.
If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,
dQ = ms dT2
The rate of loss of heat is given by,
dQ/dt = ms (dT2/dt) ……..(2)
Compare the equations (1) and (2) as,
– ms (dT2/dt) = k (T2 – T1)
Rearrange the above equation as:
dT2/(T2–T1) = – (k / ms) dt
dT2 /(T2 – T1) = – Kdt
where K = k/m s
Integrating the above expression as,
loge (T2 – T1) = – K t + c
or
T2 = T1 + C’ e–Kt
where C’ = ec