Latent Heat of Fusion and Ice Maker Problem

[Silent]

Homework Statement

You are required to calculate the efficiency of an ice-making machine that takes in water at a temperature of 16°C and produces ice cubes at a temperature of -6°C. Water is taken in at the rate of 1kg every 5 minutes and the input power to the machine is 300W. The Specific Heat Capacity of water is 4200J/kg, the specific Heat Capacity of Ice is 2100J/kg and the specific Latent Heat of Fusion is 335kJ/kg.

Homework Equations

Q = mcΔt
Q = mL_F

The Attempt at a Solution

Input energy:
Q_5mins = (300)(300) = 90 x 10³J
Energy to cool ice from 16°C to 0°C:
Q = (1)(4200)(16) = 67.2 x 10³J
Energy to change water to ice:
Q = (1)(335 x 10³) = 335 x 10³J
Energy to cool ice from 0°C to -6°C:
Q = (1)(2100)(6) = 12.6 x 10³J
Total energy required to make the ice:
Q_T = 67.2 x 10³J + 335 x 10³J + 12.6 x 10³J = 414.8 x 10³J

We haven't done any efficiency calculations at all, so I'm not sure of the process. I know (or think I do), the energy supplied by the ice cube maker and the energy required to make the ice cubes, but I'm not sure how to combine the two. My instinct is to divide one by the other, but that gives a very low efficiency rating of ~21%. All the stuff I've found on the intarwebs refers to the Carnot Cycle, is that relevant here?

Thank you to anyone that responds.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[SteamKing]

We haven't done any efficiency calculations at all, so I'm not sure of the process. I know (or think I do), the energy supplied by the ice cube maker and the energy required to make the ice cubes, but I'm not sure how to combine the two. My instinct is to divide one by the other, but that gives a very low efficiency rating of ~21%.

Were you expecting a higher efficiency for an ice maker than say, a piston engine?

The following is the definition of efficiency as it relates to mechanical devices:

"The efficiency of a heat engine relates how much useful work is output for a given amount of heat energy input."

See this article:

http://en.wikipedia.org/wiki/Heat_engine#Other_criteria_of_heat_engine_performance

All the stuff I've found on the intarwebs refers to the Carnot Cycle, is that relevant here?

Have you studied thermal cycles at all? Do you think you have anything remotely resembling a Carnot Cycle process here?

http://en.wikipedia.org/wiki/Carnot_cycle

 

[Silent]

Thank you for the reply.

Actually, yes - I was expecting a higher figure for efficiency.

Partly because an ice-maker is a so-called white good which tend to have high (advertised) ratings of efficiency, and partly because the ~21% figure is derived from dividing the 90,000J by the 414800J, which I'm fairly sure is incorrect.

Carnot's cycle does seem relevant, but it mentions hot and cold reservoirs, and those're what led me to the conclusion that it wasn't directly relevant.

 

[SteamKing]

Carnot cycles are studied because they represent the theoretical maximum efficiency which can be obtained for a thermodynamic process. All real cycles by definition are less efficient than a Carnot cycle operating between the same maximum and minimum absolute temperatures.

IDK why you are under the impression that so-called 'white goods' are somehow endowed with higher efficiencies than other mechanical devices. Advertising is not a good basis on which to form such an opinion. If you read the attached articles, you should know that any real mechanical device with an efficiency approaching 50% is pretty rare. The efficiency of an automobile engine is decidedly less than 50%, and how much advertising do you think is done to sell cars?

 

[AlephZero]

The following is the definition of efficiency as it relates to mechanical devices:

"The efficiency of a heat engine relates how much useful work is output for a given amount of heat energy input."

This device is not a heat engine. It is the opposite: a heat pump. You supply energy (in this case, electrical not mechanical) and use it to transfer heat energy from one place (the water) to another (probably the air in the room where the ice-maker is being used).

The usual term for the "efficiency" of a heat pump is "coefficient of performance" (COP) and it can be (and often is) greater than 100%. http://en.wikipedia.org/wiki/Coefficient_of_performance

 

[Silent]

Thanks for the replies.

One thing my tutor did say is '...and if you're getting something more than 100% you known it's wrong, because something can't be more than 100% efficient.'

I don't think he means generally. I think he means in this scenario. He's looking for an answer between 0 and 100%.

 

[rude man]

This device is not a heat engine. It is the opposite: a heat pump. You supply energy (in this case, electrical not mechanical)

It's still mechanical: electrical → mechanical → heat removal. The compressor is an electrical-to-mechanical transducer.

 

[Silent]

So if η = W/QT in this case, that means that the efficiency is 21.70%.

That doesn't seem right, but okay.

 

[rude man]

Homework Statement

You are required to calculate the efficiency of an ice-making machine that takes in water at a temperature of 16°C and produces ice cubes at a temperature of -6°C. Water is taken in at the rate of 1kg every 5 minutes and the input power to the machine is 300W. The Specific Heat Capacity of water is 4200J/kg, the specific Heat Capacity of Ice is 2100J/kg and the specific Latent Heat of Fusion is 335kJ/kg.

Homework Equations

Q = mcΔt
Q = mL_F

The Attempt at a Solution

Input energy:
Q_5mins = (300)(300) = 90 x 10³J
Energy to cool ice from 16°C to 0°C:
Q = (1)(4200)(16) = 67.2 x 10³J
Energy to change water to ice:
Q = (1)(335 x 10³) = 335 x 10³J
Energy to cool ice from 0°C to -6°C:
Q = (1)(2100)(6) = 12.6 x 10³J
Total energy required to make the ice:
Q_T = 67.2 x 10³J + 335 x 10³J + 12.6 x 10³J = 414.8 x 10³J

Your calculations are all correct. Now you need to understand efficiency: e = heat removed/electrical energy input'ed every 5 minutes.

So e = 414.8e3/9e4 = 4.6.

This may blow your mind sice that's 460 %, but it's correct. So this is not really an efficiency calculation but a 'coefficient of performance' (COP). The fact that 1 unit of input energy results in several units of heat removal explains the popularity of heat pumps where the temperatures in winter don't drop too far.

So when your tutor said 'anything > 100% must be wrong' he's pulling your leg.

EDIT: I previously miscalculated the reversible-cycle (i.e. max. possible) efficiency.

We can compare your cycle to a reversible cycle by comparing the work needed to do the job.

Let W = your work
W' = work of a reversible cycle

Then, setting the change in entropy of the universe to zero we get
QT/T2 = (QT + W')/T1.

We have to assumte T1 which is essentially the temperature of the condenser coil. Assume room temp. = 300K and the coil is fan-cooled to 300K also. Then

e = W'/W = QT(T1/T2 - 1)/W
& I computed e = W'/W = 0.57.

But that is more or less meaningless. Refrigerators do not operate in a reversible cycle, not even theoretically - the expansion valve (throttling) process is by definition irreversible.