Question

150 grams of boiling water (temperature 100°C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 970 grams and initial temperature 25°C (the heat capacity of aluminum is 0.9 J/gram/K). (a) After a short time, what is the temperature of the water?

(b) What simplifying assumptions did you have to make?

The thermal energy of the aluminum doesn't change.

Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.

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.

(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 24054 J of work, and the temperature of the water and pan increases to 83.4°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?

Answer #1

a)

Temp. of water = 100 + 273 = 373 K

Temp of pan = 25 + 273 = 298 K

Thermal energy of water,

Ethermal = mc*T = 150 g x 4.2 J/g/k x 373 K = 234990 K

Thermal energy of pan,

Ethermal = 970 g x 0.9 J/g/k x 298 k = 260154 K

Esystem = 234990 K + 260154 K = 495144 K

T(150 x 4.2 + 970 x 0.9) = 495144

T = 329.44 K = 56.44 C

b)

Energy transfer between system and surroundings are negligible during this time heat capacities of both water and aluminium hardlty in temp. in this range.

c)

Q = E - W

E = mw*Cv*T + mp*Cv*T

E = [150*4.2(83.4 - 56.44) + [970*0.9*(83.4 - 56.44)

E = 40520.88 J

Q = 40520.88 - 24054

Q = 16466.88 J

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