Question

How long would a 3.40 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 2.50) does when it freezes 1.60 kg of water at 18.8°C into ice at 0°C?

Answer 1

The heat needed to lower the temperature of 1.36 kg water at 19.5°C to water at 0°C is
… ΔQ₁ = m c ΔT = ( 1.40 kg ) ( 4.18 J g ⁻¹ K ⁻¹ ) ( 0 ‒ 18.8 )°C
………. = ( 1.140 kg ) ( 1000 g kg⁻¹ ) ( 4.18 J g ⁻¹ °C ⁻¹ ) ( ‒ 18.8°C )
………. = -89585.76 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 …
… ΔQ₂ = - m • Lf = - ( 1.40 kg ) (334 J / g )
………. = - ( 1.40kg ) ( 1000 g kg⁻¹ ) (334 J / g )
………. = - 4676600 J …
The total heat energy removed from the 1.36 kg water is then
… ΔQ = ΔQ₁ + ΔQ₂ = -89585.76 J - 467600 J = -557185.75 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 = 2.50, we find that …
… | Qa | / { | Qb | ‒ | Qa | } = 2.50 … --> … | Qa | = ( 2.50 ) { | Qb | ‒ | Qa | } …
… | Qa | = ( 2.50 ) | Qb | ‒ ( 2.50) | Qa | …
… ( 1 + 2.50 ) | Qa | = ( 3.50 ) | Qa | = ( 2.50 ) | Qb | … so that …
… | Qb | = ( 3.50 / 2.50 ) | Qa | = ( 1.4) | ΔQ |
……….. = ( 1.4 ) ( 557185.75 J ) = 780060.05 J … that is, with three
significant figures … | Qb | = 7.80 × 10 ⁵ J … using a 3.40 kW space heater,
we have, using the definition of power P = ΔE / Δt , the expression for the
elapsed time … Δt = ΔE / P = | Qb | / P = 7.80 × 10 ⁵ J / 3.40 kW …
… Δt = 7.79 × 10 ⁵ J / ( 3.40 × 10 ³ J / s ) = 26.52 s = 4.42 min