A 56.1 grams copper bowl contains 86.6 g of water at 26.8 °C. A very hot copper

Question

A 56.1 grams copper bowl contains 86.6 g of water at 26.8 °C. A very hot copper cylinder, with a mass of 472.1 g, is dropped into the water, causing the water to boil, with 2.06 g being converted to steam. The final temperature of the system is 100.0°C.
Assume the specific heat capacity of water is 4.184 J/g-K, and for copper is 0.386 J/g-K, and that the latent heat of boiling of water is 2256 J/g.
How much heat was transferred to the water?
How much to the bowl ?
What is the original temperature of the cylinder? (answer in °C))

Explanation / Answer

a) Heat supplied to water:

86.6x 4.184 x ( 100-26.8) + 2.06 x 2256 = 26522.87808 + 4647.36

Total heat supplied to water = 31170.23808 J

b) heat supplied to bowl =

56.1x0.386 x (100-26.8)=1585.11672 J

c)Heat given by copper cylinder = 31170.23808+ 1585.11672= 32755.3548 J

Let the initial temperature of Cu cyliunder = t

32755.3548=472.1 x 0.386 x ( t-100)

179.7467 = t - 100

t = 279.7467 C

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