200 grams of water are in a 60 gram aluminum cup (part of a well insulated calor

Question

200 grams of water are in a 60 gram aluminum cup (part of a well insulated calorimeter). A thermometer of negligible mass is immersed in the water and has read 22 degree C for the last 20 minutes. A 50 gram copper cylinder that has been in a freezer (interior temperature of -8 degree C) all night is quickly removed from the freezer and dropped into the water. Assuming that the system consisting of the water, the cup and the cylinder is closed to energy flows: What does the thermometer read after the cylinder has been in the water for a long time^1? If the cup (still containing the water and the copper cylinder) is removed from the calorimeter and allowed to warm to the room temperature (also 22 degree C) how much energy flows into the system?

Explanation / Answer

heat lost by water + aliinium = heat gained by copper

mwater*cwater*dt1 + maluminium*caluminium*dt1 = mcopper*ccopper*dt2

(0.2*4186*(22-t)) + (0.06*900*(22-t)) = 0.05*390*(t+8)

t = 21.4 degrees

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part b

heat requied by water+aluminium to raise to 22

Q = (0.2*4186*(22-21.4)) + (0.06*900*(22-21.4))

Q = 534.72 J

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