A 110 g copper bowl contains 220 g of water, both at 22.0°C. A very hot 380 g co

Question

A 110 g copper bowl contains 220 g of water, both at 22.0°C. A very hot 380 g copper cylinder is dropped into the water, causing the water to boil, with 14.1 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

(a)
Energy transferred to water, Qw = m*c*T + m * Ls
Qw = 220*4.186*(100 - 22.0) + 14.1 * 2257
Qw = 103655.46 J
Qw = 24.77 KCal

(b)
Energy transferred to Bowl, Qb = m*c*T
Qb = m*c*T
Qb = 110 * 0.386 * (100 - 22.0)
Qb = 3311.88 J
Qb = 791.56 Cal

(c)
Let the original Temp of cylinder be Ti.
Heat Energy Lost by cylinder = Heat Energy gained by Water + Bowl
380 *0.386 *(Ti - 100) = 103655.46 + 3311.88
Ti = 829.3 °C

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