A pitcher having 1-mm thick porous walls holds 10 kg of water. Water reaches the outer surface and evaporates at a rate of 0.1 g s^ -1 . The surface area of the pitcher (one side) is 200 cm^2 . The room temperature is

Question

Quantrex Academy · Accepted Answer

Correct answer: C. 28^ ^ ^ C

In steady state, the heat required for evaporation of water is supplied by the heat conducted from the surroundings through the walls of the pitcher.

The rate of heat required for evaporation is:

\dfrac{dQ}{dt} = L \dfrac{dm}{dt}

Substituting the given values:

\dfrac{dQ}{dt} = (2.27 \times 10^6 \text{ J kg}^{-1}) \times (0.1 \times 10^{-3} \text{ kg s}^{-1}) = 227 \text{ J s}^{-1}

The rate of heat conduction through the walls is given by:

\dfrac{dQ}{dt} = \dfrac{K A (T_0 - T)}{d}

Equating the two rates:

\dfrac{0.80 \times (200 \times 10^{-4}) \times (42 - T)}{10^{-3}} = 227

\dfrac{0.016}{10^{-3}} \times (42 - T) = 227

16(42 - T) = 227

42 - T = \dfrac{227}{16} = 14.1875

T = 42 - 14.1875 = 27.8125^{}^{{}^{\circ}} \text{C}

Rounding to the nearest integer, we get

T \approx 28^{}^{{}^{\circ}} \text{C}

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A pitcher having 1-mm thick porous walls holds 10 kg of water....

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