How much work must an ideal refrigerator perform to make 1 kg of ice at -10C from 1 kg of
water at 20C when 20C is also its exhaust temperature?
To determine the amount of work required by an ideal refrigerator to make 1 kg of ice at -10°C from 1 kg of water at 20°C, we can use the formula for the work done by a refrigerator:
W = Q_h - Q_c
where W is the work done, Q_h is the heat absorbed from the cold reservoir, and Q_c is the heat rejected to the hot reservoir.
In this case, we're given that the exhaust temperature of the refrigerator (the hot reservoir) is also 20°C. Since the refrigerator is ideal, it operates on the reverse Carnot cycle, meaning it is reversible and has maximum efficiency.
To calculate Q_h, we can use the equation:
Q_h = m * c * (T_h - T_c)
where m is the mass of the substance (in this case, water), c is its specific heat capacity, T_h is the hot reservoir temperature, and T_c is the cold reservoir temperature.
Given that the mass of water is 1 kg, the specific heat capacity of water is approximately 4186 J/kg°C, T_h is 20°C, and T_c is -10°C, we can calculate Q_h:
Q_h = 1 kg * 4186 J/kg°C * (20°C - (-10°C)) = 1 kg * 4186 J/kg°C * 30°C = 125,580 J
Next, to calculate Q_c, we can use the same formula, but this time T_h and T_c will be reversed:
Q_c = m * c * (T_c - T_h)
Q_c = 1 kg * 4186 J/kg°C * (-10°C - 20°C) = 1 kg * 4186 J/kg°C * (-30°C) = -125,580 J
Since the work done is the difference between the two, we can now calculate W:
W = Q_h - Q_c = 125,580 J - (-125,580 J) = 251,160 J
Therefore, the ideal refrigerator must perform 251,160 J of work to make 1 kg of ice at -10°C from 1 kg of water at 20°C when the exhaust temperature is also 20°C.