A 125 g aluminum cylinder is removed from a liquid nitrogen bath, where it has been cooled to −196° C. The cylinder is immediately placed in an insulated cup containing 1270 g of water at 15.0° C,. What is the equilibrium temperature of this system? If your answer is 0° C, determine the amount of water that has frozen. The average specific heat of aluminum over this temperature range is 653 J/(kg · K).
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Assume the final temperature of the aluminium is 0 deg C.
It will absorb
0.125 * 196 * 653 = 15998.5 J in warming to this temperature.
The specific heat of water (from Wikipedia) is 4181.3 J/(kg.K)
If the water cools to 0 deg C, then it will release
1.270 * 4181.3 * 15 = 79653.765 J
The water will not cool to freezing point.
Now let the equilibrium temperature be T deg C
1.270 * 4181.3 * (15 - T) = 0.125 * 653 * (T + 196)
5310.251 * (15 - T) = 81.625 * (T + 196)
79653.765 - 5310.251T = 81.625T + 15998.5
79653.765 - 15998.5 = (5310.251 + 81.625)T
T = 63655.265 / 5391.876 = 11.806 deg C (to 3 decimal places)
It will absorb
0.125 * 196 * 653 = 15998.5 J in warming to this temperature.
The specific heat of water (from Wikipedia) is 4181.3 J/(kg.K)
If the water cools to 0 deg C, then it will release
1.270 * 4181.3 * 15 = 79653.765 J
The water will not cool to freezing point.
Now let the equilibrium temperature be T deg C
1.270 * 4181.3 * (15 - T) = 0.125 * 653 * (T + 196)
5310.251 * (15 - T) = 81.625 * (T + 196)
79653.765 - 5310.251T = 81.625T + 15998.5
79653.765 - 15998.5 = (5310.251 + 81.625)T
T = 63655.265 / 5391.876 = 11.806 deg C (to 3 decimal places)