A 130 g copper bowl contains 190 g of water, both at 21.0°C. A very hot 460 g co

Question

A 130 g copper bowl contains 190 g of water, both at 21.0°C. A very hot 460 g copper cylinder is dropped into the water, causing the water to boil, with 15.4 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 = 100 *0.0923* [100 - 21] = 729.18 cal.
Heat gained by 190 gm of water = 190 *1*[79] =15010 cal.
Heat used to convert 15.4 g of water into steam = mL
= 539*15.4 = 8300.6 cal.
Total calories gained by water and bowl= 24039.79 cal.
--------------------------------------...
Heat lost by 460g of copper is 460* 0.0923*[ - 100]
= 42.459[ - 100] cal.
==============================
Equating the two
42.459[ - 100] = 24039.79
[ - 100] = 500.98
[] =566.18° C.
--------------------------------------...
a) 15010 + 8300.6 = 23310.6 cal.
b) 729.18 cal
c) 566.18° C.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.