A 180 g copper bowl contains 220 g of water, both at 23.0°C. A very hot 490 g co

Question

A 180 g copper bowl contains 220 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 12.7 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.

I figured out b but I can't seem to get a. I solved (m for water)(c for water)(change in temp)+(Latent heat )(mass of steam)

(220)(1)(77)+(.539)(12.7)

Whenever I do this I am getting 16946.8cal which wiley is saying is wrong.

Explanation / Answer

Let the temperature of cylinder is t*C.
(a) Heat transferred to water = m x s x delta T + m x L
=>H = 220 x 4.18 x (100 - 23) + 12.7 x 2270
=>H = 100 kJ = 86.79/4.18 = 23.836 Kcal
(b)By Q = m x s x delta T
=>Q = 180 x 0.385 x (100 - 23)
=>Q = 5.336 kJ = 4.62/4.18 = 1.276 kcal
(c) [Q+H] = m x s x delta T
=>[86790+4620] = 490 x 0.385 x (t-100)
=>t = 584.55*C

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