How long would a 2.60 kW space heater have to run to put into a kitchen the same
ID: 1913950 • Letter: H
Question
How long would a 2.60 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 3.05) does when it freezes 1.48 kg of water at 19.1Explanation / Answer
This is the method Values may differ The heat needed to lower the temperature of 1.36 kg water at 19.5°C to water at 0°C is … ?Q1 = m c ?T = ( 1.36 kg ) ( 4.18 J g ?¹ K ?¹ ) ( 0 ? 19.5 )°C ………. = ( 1.36 kg ) ( 1000 g kg?¹ ) ( 4.18 J g ?¹ °C ?¹ ) ( ? 19.5°C ) ………. = -110,853.6 J … negative since heat energy is removed … The heat needed to freeze 1.36 kg water at 0°C into ice at 0°C is … … ?Q2 = - m • Lf = - ( 1.36 kg ) (334 J / g ) ………. = - ( 1.36 kg ) ( 1000 g kg?¹ ) (334 J / g ) ………. = - 454,240 J … The total heat energy removed from the 1.36 kg water is then … ?Q = ?Q1 + ?Q2 = -110,853.6 J - 454,240 J = -565,093.6 J … According to the definition of coefficient of performance (COP) of a refrigerator … … COP = | Qa | / { | Qb | ? | Qa | } … where … | Qa | = amount of heat removed from the low temperature part … | Qb | = amount of heat given off to the high temperature part … since in this particular case COP = 2.64, we find that … … | Qa | / { | Qb | ? | Qa | } = 2.64 … --> … | Qa | = ( 2.64 ) { | Qb | ? | Qa | } … … | Qa | = ( 2.64 ) | Qb | ? ( 2.64 ) | Qa | … … ( 1 + 2.64 ) | Qa | = ( 3.64 ) | Qa | = ( 2.64 ) | Qb | … so that … … | Qb | = ( 3.64 / 2.64 ) | Qa | = ( 1.378788 ) | ?Q | ……….. = ( 1.378788 ) ( 565,093.6 J ) = 779,144 J … that is, with three significant figures … | Qb | = 7.79 × 10 5 J … using a 3.10 kW space heater, we have, using the definition of power P = ?E / ?t , the expression for the elapsed time … ?t = ?E / P = | Qb | / P = 7.79 × 10 5 J / 3.10 kW … … ?t = 7.79 × 10 5 J / ( 3.10 × 10 ³ J / s ) = 251.29 s = 4.19 min
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