A 170 g copper bowl contains 120 g of water, both at 24.0°C. A very hot 480 g co

Question

A 170 g copper bowl contains 120 g of water, both at 24.0°C. A very hot 480 g copper cylinder is dropped into the water, causing the water to boil, with 4.90 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

Cwater = specific heat = 1cal/ g*K = 4186J / kg0C

Lwater = latent heat = 539 cal / kg = 2263.8 J/ kg                                                      ( 1 cal = 4.2 J)

dT = change in Temp.

i) energy gained by water = energy for 120g of water from 240C to 1000C + energy by 4.90 g of water to convert to steam

                                     = m*Cwater*dT + m*Lwater              

                                     = 120*10-3*4186*( 100-24) + 4.90*10-3*2263.8

                                     = 38187.41 J

ii) energy gained by bowl = m*Ccopper*dT                                      ( Ccopper = 0.0923 cal/gK = 387J / kg0C)

                                = 170*10-3*387*(100-24)

                                  = 5000 J

iii) let T = initial Temp. of cylinder

total energy lost by cylinder = energy gained by water + bowl

480*10-3*387 *( T- 100) = 38187.41 + 5000

T = 332.490C

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