A body of area `0.8xx10^-2m^2` and mass `5xx10^(-4)kg` directly faces the sun on a clear day. The body has an emissivity of 0.8 and specific heat of `0.8 cal//kg` K. The surroundings are at `27^@C`. (solar constant `=1.4 kW//m^(2)`). Q. The rate of rise of the body's temperature is nearly

Options

A. `0.36^@C//s`
B. `3.6 K//s`
C. `36^@ C//s`
D. `72 K//s`

Answer

B

Explanation

We consider the leaf to be a black body. The rate of energy radiated at any instant.
`((dQ)/(dt))_e=sigmaeAT_0^4` .(i)
If m is the mass of the body and C is its specific heat then the heat gained
`(dQ)/(dt)=mc(dT)/(dt)` .(ii)
Since intensity of sun beam is uniform power incident on the body is SA heat absorbed `((dQ)/(dt))_(ab)=SAe`
Rate of increase of temperature `=((n et power ab so rbed))/(thermal ca pacity of body)`
`=((SAalpha-sigmaAeT^3))/(mc)`
Given
`sigma=5.67xx10^-8(J)/(sm^2K^4)`
`A=0.8xx10^-2m^2`,`alpha=e=0.8`
`S=1.4xx10^3(W)/(m^2)`
`T=300K`
`m=5xx10^-4kg`
`C=0.8(kcal)/(kg)K=0.8xx4.2(kJ)/(kgK)`
`(dT)/(dt)=([1.4xx10^3xx0.8xx10^-2xx0.8-5.67xx10^-8xx0.8xx10^-2xx0.8xx(300)^4])/(5xx10^-4xx(0.8xx4.2xx10^3))`
`=((8.96-2.9))/(1.68)=(3.6^@C)/(s)=3.6(K)/(s)`