problem 33: Suppose 525.0 J of heat is added to a 111-g piece of aluminum at 20.

Question

problem 33: Suppose 525.0 J of heat is added to a 111-g piece of aluminum at 20.0° C. What is the final temperature of the aluminum?
problem 49: A glass window 0.35 cm thick measures 85 cm by 36 cm. How much heat flows through this window per minute if the inside and outside temperatures differ by 13° C? problem 33: Suppose 525.0 J of heat is added to a 111-g piece of aluminum at 20.0° C. What is the final temperature of the aluminum? problem 49: A glass window 0.35 cm thick measures 85 cm by 36 cm. How much heat flows through this window per minute if the inside and outside temperatures differ by 13° C?

Explanation / Answer

(a) Heat Q = 525 J mass m = 111g = 0.111 Kg Initial temperature t1 = 20° C Final temperature t2 = ? specific heat of aluminum S = 910 J /Kg ° C We know that Q = ms(t2 - t1) 525 = 0.111*910*(t2 - 20) t2 = 20 + 5.197     = 25.197° C (b) Area A = 0.2975 m^2 thickness of glass d = 0.35 cm = 0.35*10^-2 m coefficient of thermal conductuivity K = 1.05 W/mK temperature difference T = 13 + 273 = 286 K time t = 60 s heat flows through this window per minute Q = KATt/d     = (1.05*0.2975*286*60)/(0.35*10^-2)         = 1531530 J (a) Heat Q = 525 J mass m = 111g = 0.111 Kg Initial temperature t1 = 20° C Final temperature t2 = ? specific heat of aluminum S = 910 J /Kg ° C We know that Q = ms(t2 - t1) 525 = 0.111*910*(t2 - 20) t2 = 20 + 5.197     = 25.197° C (b) Area A = 0.2975 m^2 thickness of glass d = 0.35 cm = 0.35*10^-2 m coefficient of thermal conductuivity K = 1.05 W/mK temperature difference T = 13 + 273 = 286 K time t = 60 s heat flows through this window per minute Q = KATt/d     = (1.05*0.2975*286*60)/(0.35*10^-2)         = 1531530 J

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