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 0.05m2. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in ^\circ C) in the temperature of water and the surroundings after a long time will be \underline\hspace2cm . (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)

Heat and Thermodynamics PYQ — JEE Advanced 2020 · Shift 2 · Q54 with Solution

View written solutionFree

Correct answer: 8.33

The solar intensity is given as $I = 700\ \text{W m}^{-2}$ and the effective absorbing area is $A = 0.05\ \text{m}^2$

So, the rate of heat gain by water is $P_{\text{in}} = IA = 700 \times 0.05 = 35\ \text{W}$

Let the temperature difference between water and surroundings after a long time be $\Delta T = T - T_s$

Newton's law of cooling here is given with constant $k = 0.001\ \text{s}^{-1}$

For a body of mass $m$ and heat capacity $C$ (specific heat), the cooling equation is $mC\frac{dT}{dt} = -mCk(T-T_s)$

Thus, the rate of heat loss is $P_{\text{loss}} = mCk\Delta T$

Given:

So, $P_{\text{loss}} = 1 \times 4200 \times 0.001 \times \Delta T = 4.2\Delta T$

After a long time, temperature becomes constant, so heat gained equals heat lost: $P_{\text{in}} = P_{\text{loss}}$

Therefore, $35 = 4.2\Delta T$

Hence, $\Delta T = \frac{35}{4.2} = 8.33\ \text{K}$

Since temperature difference in kelvin and degree Celsius are numerically equal, $\Delta T = 8.33^\circ \text{C}$

The required temperature difference is $\boxed{8.33}$