Explanation Draw the schematic diagram of the room as shown in figure (1). Write the expression for the volume of the room: V = l b h (I) Here, length of the room is l , width of the room is b , and height of the room is h . Write the formula for the mass of the air. m a i r = P 1 V 1 R T 1 (II) Here, pressure is P 1 , volume is V 1 , gas constant is R , Temperature is T 1 . Write the expression for the change in internal energy of the water. Δ U w a t e r = m c P ( T 2 − T 1 ) (III) Here, mass of the water is m , specific heat at constant pressure is c P , final temperature of the water is T 2 , and temperature of the water is T 1 . Write the expression for the change in internal energy of the air. Δ U a i r = m c v ( T 2 − T 1 ) a i r (IV) Here, mass of the air is m , specific heat at constant volume of the air is c v , final temperature of the air is T 2 , and temperature of the air is T 1 . Express the energy balance equation for closed system. E i n − E o u t = Δ E s y s t e m 0 = ( Δ U ) w a t e r + ( Δ U ) a i r (V) Here, energy transfer in to the control volume is E i n , energy exit from the control volume is E o u t and change in internal energy of system is Δ E s y s t e m . Conclusion: Substitute 4 m for l , 5 m for b , and 6 m for h in equation (I). V = 4 m × 5 m × 6 m =120 m 3 Substitute 95 kPa for P 1 , 120 m 3 for V 1 , 0.2870 kPa ⋅ m 3 kg ⋅ K for R , 15 ° C for T 1 in equation (II) m a i r = 95 kPa × 120 m 3 0

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ISBN: 9780100257054

Author: CENGEL

Publisher: YUZU

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Chapter 4.5, Problem 122RP

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The final equilibrium temperature in the room

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Chapter 4.5, Problem 121RP

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