9.A refrigerator can convert 5 points 30 g of water at 22 'C to 3O g of ice at 20 C per 60 second. If COP of the refrigerator is 3 and room temperature is 22 C while its ice tray compartment is at 20 C,then calculate the input in watts. *
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The heat absorbed by the refrigerator can be calculated using the formula: Q_absorbed = m * c * ΔT where: m = mass of water = 30 g c = specific heat capacity of water = 4.18 J/g°C ΔT = change in temperature = (22 - 20) °C = 2 °C Q_absorbed = 30 g * 4.18 J/g°C * 2 Show more…
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A refrigerator has a coefficient of performance (COP) of $3.00$. The ice tray compartment is at $-20.0^{\circ} \mathrm{C}$, and the room temperature is $22.0^{\circ} \mathrm{C}$. The refrigerator can convert $30.0 \mathrm{~g}$ of water at $22.0^{\circ} \mathrm{C}$ to $30.0 \mathrm{~g}$ of ice at $-20.0^{\circ} \mathrm{C}$ each minute. What input power is required? Give your answer in watts.
A refrigerator has a coefficient of performance of $3.00 .$ The ice tray compartment is at $-20.0^{\circ} \mathrm{C},$ and the room temperature is $22.0^{\circ} \mathrm{C}$ . The refrigerator can convert 30.0 $\mathrm{g}$ of water at $22.0^{\circ} \mathrm{C}$ to 30.0 $\mathrm{g}$ of ice at $-20.0^{\circ} \mathrm{C}$ each minute What input power is required? Give your answer in watts.
A refrigerator has a coefficient of performance of $3.00 .$ The ice tray compartment is at $-20.0^{\circ} \mathrm{C}$, and the room temperature is $22.0^{\circ} \mathrm{C} .$ The refrigerator can convert $30.0 \mathrm{g}$ of water at $22.0^{\circ} \mathrm{C}$ to $30.0 \mathrm{g}$ of ice at $-20.0^{\circ} \mathrm{C}$ each minute. What input power is required? Give your answer in watts.
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