Question

Please submit the answer to this problem on a separate piece of paper, in submission box. A copper block is removed from a 300°C oven and dropped into 1.00 kg of water at 20.0°C, which is located in an aluminum container of a mass of 0.250 kg. The system quickly reaches 25.5°, and then remains at that temperature. ho a) What is the mass of the copper block? b) How much energy did the copper block lose? c) How much energy did the water gain? d) As the temperature of water rises, the temperature of the copper block decreases. What is the temperature of the copper block at the instant that the temperature of water is 22.0°C? Hint: you need to look up all the specific heats relevant to this problem - they are in your textbook.

Answer 1

Let the mass of copper block be $M_c$

Initial temperature of copper block is $T_c = 300 \degree C$

Specific Heat of copper is $c_c = 376.8 \ J/kg \degree C$

Mass of water is

Mw = 1 kg

Initial temperature of water is $T_w = 20 \degree C$

Specific Heat of water is $c_w = 4200 \ J/kg \degree C$

Mass of aluminum container is

Initial temperature of aluminum container is $T_a = 20 \degree C$

Specific Heat of aluminum is $c_a = 921.1 \ J/kg \degree C$

Final temperature of the system is $T = 25.5 \degree C$

(a)

By principle of calorimetry,

Heat lost by the hot body = Heat gained by the cold body

Therefore, we have

$\\M_c * c_c * (T_c - T) = M_w * c_w * (T - T_w) + M_a * c_a * (T - T_a) \\\\M_c * 376.8 * (300 - 25.5) = 1 * 4200 * (25.5 - 20) + 0.25 * 921.1 * (25.5 - 20) \\\\M_c * 103431.6 = 23100 + 1266.5 \\\\M_c = \frac{24366.5}{103431.6} \\\\M_c = 0.235 \ kg$

(b)

Energy lost by the copper block is

$\\Q_c = M_c * c_c * (T_c - T) \\\\Q_c = 0.235 * 376.8 * (300 - 25.5) \\\\Q_c = 25857.9 \ J$

(c)

Energy gained by the water is

$\\Q_w = M_w * c_w * (T - T_w) \\\\Q_w = 1 * 4200 * (25.5 - 20) \\\\Q_w = 23100 \ J$

(d)

The total heat gained by water and aluminum when temperature is $22 \degree C$ , is

$\\Q = M_w * c_w * (22 - 20) + M_a * c_a * (22 - 20) \\\\Q = 1 * 4200 * 2 + 0.25 * 921.1 * 2 \\\\Q = 8400 + 460.55 \\\\Q = 8860.55 \ J$

This is the energy lost by copper block when its temperature is

therefore, we have

$\\Q = M_c * c_c * (300 - T') \\\\8860.55 = 0.235 * 376.8 * (300 - T') \\\\300 - T' = \frac{8860.55}{0.235 * 376.8} \\\\300 - T' = 100.1 \\\\T' = 300 - 100.1 \\\\T' = 199.9 \degree C$