- #1

C.Orio

- 7

- 0

## Homework Statement

In a 30% efficient, -16°C fridge, how much electrical energy will be

**used up**to lower the temperature of 250 mL of 75°C water? If the fridge is rated at 150 W, how long will this process take?

Specific Heat Capacity of Water = 4200 J/Kg×K

Specific Heat Capacity of Ice = 2000 J/Kg×K

Latent Heat of Fusion for Water = 333000 J/Kg

## Homework Equations

Q=mcΔT

Q

_{F}=mL

_{F}

Efficiency=EnergyOutput/EnergyInput × 100%

Power=Work/Δt

## The Attempt at a Solution

I've found the Q

_{Total}(Q

_{1}, Q

_{2}and Q

_{F}), which is -3500J, but I'm not sure what to do with that info, or even what it means in regards to the question. The wording is also quite confusing, and I'm not sure whether

*used up*means the input or the output. Any help would be greatly appreciated!