170 grams of bolling water (temperature 100* C, heat capacity 4.2 3/gram/K) are

Question

170 grams of bolling water (temperature 100* C, heat capacity 4.2 3/gram/K) are poured into arn C, heat capacity 4.2 3/gram/K) a heat capacity of aluminum is 0.9 3/gram/x). (a) After a short time, what is the temperature of the water? inal - 58.6 What simplifying assumptions did you have to make? The heat capacities for both water and aluminum hardly change with temperature in this temperature range The thermal energy of the water doesn't change. a Energy tranafer between the system (water plus pan) and the surroundings was negligible during this time The thermal energy of the aluminum doesn't change. (C) Next you place the pan on a hot electric stove, while the stove is heating the pan, you use a beater to stir the water, doing 27304' work, and the temperature of te water and pan increases to 82.8 C.How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan? -39139 s Additional Materials L Section 7 s My Notes Ask Your Suppose you warm up 500 grams of water (about haif a liter, or about a pint) on a stove, and while this is happening, you also stir the water with a beater doing ax1of 3 of

Explanation / Answer

(a) heat lost by the water = heat gained by the aluminium pan

m1c1(T1 - T) = m2c2(T - T2)

where m1 = 170grm, c1 = 4.2 J/g/K, T1 = 1000C, m2 = 1000grm, c2 = 0.9 J/grm/K, T2 = 210C

170 x 4.2 x (100 -T ) = 1000 x 0.9 x ( T - 21)

71400 - 714T = 900T - 18900

1614T = 71400 - 18900 = 52500

T = 52500/1614 = 32.530C Ans

(b) Answers are correct

(c) Now the final temperature of the pan +water over the stove = 82.80C

hence Q = m1c1(82.8 - 32.5) + m2c2(82.8 - 32.5)

= (170 x 4.2 + 1000 x 0.9) x 50.3

= (714 + 900) x 50.3

= 81184.2 J

but energy supplied = 27304J

hence energy supplied due to temperature difference = 81184.2 - 27304 = 53880.2J Ans

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