A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is and it is absorbed by the water over an effective area of

Question

A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is $700 {Wm}^{- 2}$ and it is absorbed by the water over an effective area of $0.05 m^{2}$ . Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in ${}^{o}C$ ) in the temperature of water and the surroundings after a long time will be _______.

(Ignore effect of the container, and take constant for Newton's law of cooling $= 0.001 s^{- 1}$ , Heat capacity of water $= 4200 J {kg}^{- 1} K^{- 1}$ ) [2020]

Smart Achievers · Accepted Answer

(8.33)

$Rate of loss of heat,$

$\frac{d Q}{d t} = σ e A (T^{4} - T_{0}^{4}) ...(i)$

$\Rightarrow \frac{d Q}{A d t} = e σ [{(T_{0} + Δ T)}^{4} - T_{0}^{4}] = σ e T_{0}^{4} [{(1 + \frac{Δ T}{T_{0}})}^{4} - 1]$

$= e σ T_{0}^{4} [(1 + 4 \frac{Δ T}{T_{0}}) - 1]$

$\frac{d Q}{A d t} = σ e T_{0}^{3} \cdot 4 Δ T ...(ii)$

Now from eq. (i)

$m s \frac{d T}{d t} = σ e A (T^{4} - T_{0}^{4})$ $[∵ Q = m s Δ T]$

$\Rightarrow \frac{d T}{d t} = \frac{σ e A}{m s} [{(T_{0} + Δ T)}^{4} - T_{0}^{4}]$

$= \frac{σ e A}{m s} T_{0}^{4} [{(1 + \frac{Δ T}{T_{0}})}^{4} - 1]$

$\frac{d T}{d t} = \frac{σ e A}{m s} T_{0}^{3} \cdot 4 Δ T$

$\Rightarrow \frac{d T}{d t} = K Δ T;$ $(K = \frac{4 σ e A T_{0}^{3}}{m s} Constant for Newton's law of cooling)$

$\Rightarrow 4 σ e A T_{0}^{3} = \frac{K}{A} (m s)$

From eq. (i)

$\frac{d Q}{A d t} = e σ T_{0}^{3} \cdot 4 Δ T$

Since, rate of loss of heat = heat received per second

$700 = (\frac{K}{A}) (m s) Δ T$ $[K \times m s = 4200 \times 10^{- 3}]$

$\Rightarrow Δ T = \frac{700 \times A}{K \times m s} = \frac{700 \times 5 \times 10^{- 2}}{10^{- 3} \times 4200} = \frac{50}{6} = \frac{25}{3}$

$∴$ $Δ T = 8.33$

Solution

Want Us to Email you About Special Offers & Updates?