A 110 g copper bowl contains 170 g of water, both at 23.0°C. A very hot 370 g co

Question

A 110 g copper bowl contains 170 g of water, both at 23.0°C. A very hot 370 g copper cylinder is dropped into the water, causing the water to boil, with 13.9 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature of the cylinder? The specific heat of water is 1 cal/g·K, and of copper is 0.0923 cal/g·K. The latent heat of vaporization of water is 539 Cal/kg.

Explanation / Answer

In this problem, water and the copper bowl will gain heat energy, while the copper cylinder will lose heat energy. •Within a phase (e.g., while water is still a liquid, or for changes in temperature of the copper cylinder or bowl), we can find the heat lost or gained from mc(T), where m is the mass, c is the specific heat, and T is the change in temperature. The specific heat of copper is cCu = 385 J/(kg.C), whereas the specific heat of water is, as usual, cw = 4186 J/(kg.C).
• If there is a phase change, the heat lost or gained can be determined from mL, where m is the mass, and L is the latent heat. The latent heat of vaporization (water to steam)is given in as Lw = 22.6 × 105 J/kg.

Since the water boils, with 13.9 g being converted into steam, we can find the energy gained by water by adding the energy required to take all 170 g of the water from 23C to 100C, and the energy required to convert 13.9 g of water to steam at 100C. So
Energy gained by water = (0.170 kg) cwater (100 23) ....{all water from 20C to 100C} +(13.9 × 103 kg)Lw..{5g of water to steam}
= 0.170* 4186 *77 + 13.9 × 103* 22.6 × 105    = 86,208.74J

Since the bowl remains solid throughout, the energy transferred to it can be found easily.

Energy gained by bowl = (0.110 kg) cCu (100 23) = 3260.95 J
The total energy lost by the cylinder is equal to the sum of the energy gained by the water and the bowl (assuming no energy was lost to the environment).

This allows us to find the initial temperature Ti of the cylinder.
0.370 kg* 385 *(Ti 100).....{Energy lost by cylinder} = 86,208.74 + 3260.95
Upon carrying out the calculation, we will find that the initial temperature of the cylinder
is Ti = 728C

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