110 grams of boiling water (temperature 100° C, heat capacity 123/gram/K) are po
ID: 1789948 • Letter: 1
Question
110 grams of boiling water (temperature 100° C, heat capacity 123/gram/K) are poured into an aluminum pan whose mass is 900 grams and initial temperature 25 C (the heat capacity of aluminum is 0.9 /gram/K) (a) After a short time, what is the temperature of the water? Tinal (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 negllgible during this time. The heat capacities for both water and aluminum hardly change with temperature in this temperature range d 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 25471 of work, and the temperature of the water and pan increases to 79.3 C How much energy transfer due to a temperature difference was there from the stove Into the system conslsting of the water plus the pan? 0Explanation / Answer
Given,
m = 110 g ; Tw = 100 deg C; Cw = 4.2 J/g K ; mAl = 900 g ; T = 25 deg
T = 100 + 273 = 373 K
Qw = M c delta T
Q = 110 x 4.2 x 373 = 172326 J
TAL = 25 + 273 = 298 K
QAL = 900 x 0.9 x 298 = 241380 J
Qtotal = 241380 + 172326 = 413706 J
Q = (mw Cw + mAl CAL) T
T = Qtotal/(mw Cw + mAl CAL)
T = 413706 / (110 x 4.2 + 900 x 0.9) = 325.24 K
T = 325.24 - 273 = 52.24 deg C
Hence, T = 52.24 deg C
b)Heat transfer between the system(water plus pan) and the surroundings was negligible during this time.
c)E = W + Q
Q = E - W
W = 25471 J
E = mW C delta T + mAl CAL TAL
E = 110 x 4.2 x (100 - 52.24) + 900 x 0.9 x (100 - 52.24) = 60750.72 J
Q = 60750.72 - 25471 = 35279.72J
Hence, Q = 35279.72 J
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