Explanation Given data: M = 5000 kg and P = 2500 W . Formula used: Write the another expression for heat energy needed to boil ( Q ) . Q = M L v (1) Here, M is the mass, and L v is the heat of vaporization. Write the expression for heat loss ( P ) . P = Q Δ t (2) Here, Δ t is the change in time. Explanation: Substitute 2500 W for P in equation (2), Q Δ t = 2500 W = 2500 J s { ∵ 1 W = 1 J s } Refer TABLE 12.5 in the textbook for Melting and boiling temperature and heats of transformation at standard atmospheric pressure that is at skin temperature of 30 ° C the heat of vaporization of water is L v = 2.4 × 10 6 J kg . Rearrange equation (1). M = Q L v Find kilograms of water must women lose to perspiration each hour ( M Δ t ) . M Δ t = ( Q L v ) Δ t = ( Q Δ t ) L v Substitute 2500 J s for Q Δ t and 2

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Chapter 12, Problem 95GP

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To find: How many kilograms of water per hour must be evaporated from an elephant’s skin to keep it at a constant temperature.

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Chapter 12, Problem 94GP

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Chapter 12 Solutions

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How many kilograms of water per hour must be evaporated from an elephant’s skin to keep it at a constant temperature.

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Chapter 12 Solutions