If the temperature of the body is increased by $ 10\%, $ the percentage increase in the emitted indication will be
- $ 46\% $
- $ 40\% $
- $ 30\% $
- $ 80\% $
The Correct Option is A
Solution and Explanation
If $T_{1}=T$ then $T_{2}=T+\frac{10}{100} T=1.1 \,T$
$\frac{Q_{1}}{Q_{2}}=\left(\frac{T}{1.1 T}\right)^{4} $
$\Rightarrow Q_{2}=1.46 Q_{1}$
$\%$ increase in energy
$=\left(\frac{Q_{2}-Q_{1}}{Q_{1}}\right)$
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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