A 170 g copper bowl contains 170 g of water, both at 24.0°C. A very hot 400 g co

Question

A 170 g copper bowl contains 170 g of water, both at 24.0°C. A very hot 400 g copper cylinder is dropped into the water, causing the water to boil, with 5.14 g being converted to steam. The final temperature of the system is 100°C. Neglect energy transfers with the environment. (a) How much energy (in calories) is transferred to the water as heat? (b) How much to the bowl? (c) What is the original temperature (in Celsius) 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

Heat gained by bowl = 170 *0.0923* [170 - 24] = 2290.886 cal.
Heat gained by 170 gm of water = 170*1*[78] =13260 cal.
Heat used to convert 3.5 g of water into steam = mL
= 539*3.5 = 1886.5 cal.
Total calories gained by water and bowl= 17437.386 cal.
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Heat lost by 400g of copper is 440* 0.0923*[ - 170]
= 40.612[ - 170] cal.
==============================
Equating the two
40.612[ - 170] = 17437.386
[ - 170] = 500.98
[] = 429.36° C.
--------------------------------------…
a) 13260+ 1886.5 = 15146.5 cal.
b) 2290.886cal
c) 429.36° C.
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