1. A 150 g copper bowl contains 250 g of water, both at 21.0°C. A very hot 430 g
ID: 1589665 • Letter: 1
Question
1. A 150 g copper bowl contains 250 g of water, both at 21.0°C. A very hot 430 g copper cylinder is dropped into the water, causing the water to boil, with 7.56 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
Use the formula mct.
Heat gained by bowl = 150 *0.0923* [100 - 21] = 1093.755 cal.
Heat gained by 250 gm of water = 250 *1*[100-21] =19750 cal.
Heat used to convert 7.56 g of water into steam = mL
= 539*7.56 = 4074.84cal.
Total calories gained by water and bowl= 24918.595cal.
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Heat lost by 430g of copper is 430* 0.0923*[ - 100]
= 39.689[ - 100] cal.
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Equating the two
39.689[ - 100] = 24918.595
[ - 100] = 627.84
[] = 727.840 C.
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a) 19750+ 4074.84 = 23824.84 cal.
b) 24918.595-23824.84=1093.755 cal
c) 727.84° C.
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