Another heat engine efficiency problem

[kbyws37]

A reversible heat engine has an efficiency of 34.2%, removing heat from a hot reservoir and rejecting heat to a cold reservoir at 0°C. If the engine now operates in reverse, how long would it take to freeze 1.10 kg of water at 0°C, if it operates on a power of 183 W?



I am getting stuck on how I can incorporate the kg of water and the power.
I don't know which equation to use that will let me use those constants and allow me to find time.

 

[hage567]

Well, to freeze water, you have to remove thermal energy from it. How do you find the amount of energy needed to be removed for a phase change of the water at 0°C to ice at 0°C?
It gives you the power the engine is operating at. What is the definition of power?
I'd think about those things to start.

 

[m2003]

I would suggest using the equation for the efficiency of a heat engine, which is efficiency = (work output/heat input) x 100%. In this case, the work output would be equal to the power of 183 W, and the heat input would be the amount of heat removed from the hot reservoir. This can be calculated by using the specific heat of water (4.186 J/g°C) and the mass of water (1.10 kg) to find the change in temperature (ΔT).

Once the heat input is determined, the equation for efficiency can be rearranged to solve for the time it would take to freeze the water, which would be the work output (183 W) multiplied by the efficiency (34.2%) and divided by the heat input (calculated using the specific heat and mass of water). This would result in a time in seconds, which can then be converted to minutes or hours as needed.

It is important to note that this calculation assumes ideal conditions and does not take into account any external factors that may affect the freezing process. Additionally, the actual time to freeze the water may vary depending on the starting temperature of the water and the efficiency of the heat engine.