Answer to Question #193534 in Molecular Physics | Thermodynamics for Tobias Amukwiita

Question

freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at 25.0°C to

1.80 kg of ice at -5.0°C in one hour. (a) What amount of heat must be removed from the water at 25.0°C

to convert it to ice at -5.0°C? (b) How much electrical energy is consumed by the freezer during this

hour? (c) How much wasted heat is delivered to the room in which the freezer sits?

Answer 1

$COP = 2.40 \\ m_i = 1.80kg\\ T_1= 25.0°C \\ T_2 = -5.0°C\\ t = 1\ hr = 3600\ sec$

$(a)\ Q = mc∆T_1+ mL + mc∆T_2\\ Q = m(c∆T_1+ L + c∆T_2)\\\\ Q = 1.80(4190(25) + 334000 + 4190(5))= 808000\ J$

$808\ kJ$ of heat must be removed from the water.

$(b)\ COP = \dfrac{Q_c}W$

$2.40 = \dfrac{808}{W}$

$W= 337\ kJ$

$(c)\ Q_w= W+ Q_c\\ Q_w = 337 + 808=1145k J$

$\therefore 1145\ kJ$ wasted heat is delivered to the room in which the freezer sits