A 170 g copper bowl contains 180 g of water, both at 23.0°C. A very hot 490 g co

Question

A 170 g copper bowl contains 180 g of water, both at 23.0°C. A very hot 490 g copper cylinder is dropped into the water, causing the water to boil, with 14.7 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. How much energy (in calories) is transferred to the water as heat?

Explanation / Answer

The heat lost of gained by a system is calculated using: dQ=m*c*dT where dQ is the heat lost of gained, m is the mass of the object or system, c is the specific heat of the object or system and dT is the change in temperature of the object or system. Since the final temperature of the system is 100oC, the heat gained by the water will be: dQ(water)=m*c*dT=170g*1.0cal/g*oC*[100oC-25oC]=170*1*75=12,750cal The heat gained by the copper bowl [specific heat of copper c=0.093cal/g*oC] will be: dQ(copper)=m*v*dT=180g*0.093cal/g*oC*[100oC-25oC]=180*.093*75=1256cal As you noted there is quite a bit of excess information in the problem as given although I can easily imagine other questions being asked about the system. (a) By Q = ms?t + mL =>Q = 180 x 10^-3 x 4187 x (100 - 20) + 180 x 10^-3 x 2270000 =>Q = 468892.80 J (b) By Q = ms?t =>Q = 170 x 10^-3 x 385 x (100 - 20) =>Q = 5236 J (c) By Q(lost) = Q(gain) =>m x s x ?t = 468892.80 + 5236 =>300 x 10^-3 x 385 x (t - 100) = 474128.8 =>t = 4205.01*C But the melting point of copper is 1086.44*C

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