A 160 g copper bowl contains185 g of water, both at 20.0°C. Avery hot 300 g copp

Question

A 160 g copper bowl contains185 g of water, both at 20.0°C. Avery hot 300 g copper cylinder is dropped into the water, causingthe water to boil, with 5.30 g beingconverted to steam. The final temperature of the system is100°C. Neglect energy transfers with the environment. (a) How much energy (in calories) istransferred to the water as heat?
1 kcal

(b) How much energy (in calories) is transferred to the bowl?
2 kcal

(c) What is the original temperature of the cylinder?
3                                  °C

I have the first two correct I just need help with the lastone.
(a) How much energy (in calories) istransferred to the water as heat?
1 kcal

(b) How much energy (in calories) is transferred to the bowl?
2 kcal

(c) What is the original temperature of the cylinder?
3                                  °C

I have the first two correct I just need help with the lastone.

Explanation / Answer

Process : Here copper cylider will lose heat and water with copperbowl will gain there heat energy From which 5.3g water will become steam at 100 degree C Let initial Temprature of copper cylinder be T we know that, Heat ,Q = mST where m = mass of material S = specific heat of that material T = change in temperature of that material For copper Cylinder Q = mST = 300*0.092*(T - 100) cal = 27.6(T - 100) For copper bowl Q = 160*0.092(100 - 20) = 1177.6 cal For water, Q = mSwT + m'L where, Sw = specific heat of water L = latent heat of vapourization => Q = 185*1*(100 - 20) + 5.3*597 = 14800 + 3164.1 =17964.1 cal By principle of calorimetry Heat loss = Heat Gain => 27.6(T - 100) = 1177.6 + 17964.1 = 19141.7 cal => T -100 = 693.54 cal => T = 793.540 C Heat loss by Copper cylinder = 19141.7 cal = 19.1417 kcal

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