A container with $1 k g$ 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 $700 W m^{- 2}$ 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 $^{\circ} C$ ) 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 $= 4200 J k g^{- 1} K^{- 1}$ )

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Solution

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

For small temperature change.

$\frac{d Q}{d t} = e σ A T^{3} Δ T$ _(i)

$\frac{m c d T}{d t} = e σ A T^{3} Δ T$

$\frac{d T}{d t} = \frac{e σ A T^{3}}{m C} Δ T$

$\frac{e σ A T^{3}}{m C} \to$ constant for Newton law of cooling

$\frac{e σ A T^{3}}{m C} = 0.001$

$e σ A T^{3} = m C \times 0.001$

$= 1 \times 4200 \times 0.001$

$e σ A T^{3} = 4.2$ _(ii)

$\frac{d Q}{d t} = 700 \times 0.05 = 35 W a t t s$ ___(iii)

Put the value of equation (ii), and equation (iii)
in equation (i)

$35 = 4.2 Δ T$

$\frac{35}{4.2} = Δ T$

$Δ T = 8.33$

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Q. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for is taken from the water itself and the water is cooled down. Assume that a pitcher contains

10 k g

of water and

0.2 g

of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by

5^{o} C

. Specific heat capacity of water

{= 4200 J {k g}^{- 1}} C^{- 1}

and latent heat of vaporization of water

= 2 \cdot 27 \times 10^{6} J {k g}^{- 1}

Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by $5^{\circ} C$ . Specific heat capacity of water $= 4200 J k g^{- 1}$ $^{\circ} C^{- 1}$ and latent heat of vaporization of water $= 2.27 \times 10^{6} J k g^{- 1}$