A 140 g copper bowl contains 190 g of water, both at 21.0°C. A very hot 450 g copper cylinder is dropped into the water, causing the water to boil, with 11.8 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.
Q = cm(Tf - Ti)
Q = Lm
The Attempt at a Solution
Qw = (1 cal/g*C)(190g)(100 C - 21 C) = 15,010 cal
Qb = (.0923 cal/g*C)(140g)(100 C - 21 C) = 1020.8 cal
Qc = (.0923 cal/g*C)(450g)(100 C - Ti) = 4153.5 - 41.5
Lm = (539 cal/g)(11.8g) = 6360.2 cal
Qc = Qw + Qb + Lm
4153.5 - 41.5Ti = 15,010 + 1020.8 + 6360.2
Ti = -439.46 C
The energy transferred in this problem is just Q right? So the energy transferred to the bowl would be what Qb is equal to? My final answer for the cylinders initial temperature is clearly wrong. What did I screw up in my work? Isn't it Q lost = Q gained?