Thermodynamics- Refrigeration of water/ice

[Murgs2012]

1. Homework Statement
A refrigerator is to freeze 150g of water at 0°c in one minute, the ambient temperature being 20°C. Calculate in Watts the minimum power required?

So known stuff:
mass of water =0.15kg
Temp of cold reservoir= 273k
Temp of warm reservoir= 293k
Specific heat capacity of water = 4190 J/kg k
Specific latent heat of fusion of water= 3.33x105 J/kg

2. Homework Equations
Q=mcΔt
Q=±mLf
COP= Q2/W
W=Q1-Q2
Power=W/T
Q1/Q2 = T1/T2
3. The Attempt at a Solution

First off i thought good idea to work out how much energy would take to actually freeze all the water so which was like (0.15*4190*-20)+(0.15*-333000)= -62520 J (assume that means that 62520J was released to the system from the water as it converted to ice)

I thought this was the Q1 value, then knowing T1 was 273 k and T2 was 293 k subbed into Q1/Q2 = T1/T2 to solve for Q2 which was (-62520)(293)/ (273) which gets Q2=-67000.

But then Work done= Q1-Q2 so like -62520--67000= 4580J

Then work done/time = power

and then like 4580/60 = 76 Watts... but that's too small to freeze water that fast surely?


Any ideas?

 

[Murgs2012]

Delete i wrote it down wrong.

 

[Murgs2012]

heres my next go at it...
1. Homework Statement
A refrigerator is to freeze 150g of water at 0°c in one minute, the ambient temperature being 20°C. Calculate in Watts the minimum power required?

So known stuff:
mass of water =0.15kg
Temp of cold reservoir= 273k
Temp of warm reservoir= 293k
Specific latent heat of fusion of water= 3.33x105 J/kg

2. Homework Equations

Q=±mLf
COP= QH/W
W=QH+QC
Power=wd/T
Q1/Q2 = -T1/T2
3. The Attempt at a Solution

First of, i assume the water is already at 0°c or 273k, so need only need to calculate the Latent heat of fusion for the water. As the water is freezing i think its Q=-mLf so -(0.15x3.33x105) = -49950J.

I then said this was the value for Qc.
The question asks for the minimum power required so i though we assume its a carnot engine, so max efficiency, so
Qc/Qh= -Tc/Th to calculate the value for Qh. which gets (-49950*293)/(273) which gets a Qh value of 53609J.

So seeing as work done on the system is equal to Qh+Qc then work done would be 53609+(-49950) J
=3659J work done on engine.

then power=work done/ time
so 3659/60 = 61 Watts.

i know its assuming that the system is max efficient but that just feels low.

 

[rude man]

You were right the first time! (I did not check your computations on the total heat removed from the water though).

EDIT: I see now you did miscalculate the heat removed from the water. It's just the latent heat of fusion that's to be removed.

And 61W is the correct answer, assuming you computed the latent heat correctly.

 

[rezeew]

Your calculations seem to be on the right track, but there are a few things that need to be clarified. First, the specific heat capacity of water is usually given as 4.18 J/g°C, not 4190 J/kg k. This may be where your calculations went off.

Also, when calculating the work done, you need to use the absolute value of Q2, since work is a scalar quantity and cannot be negative. This means that Q2 should be 67000 J, not -67000 J.

Finally, it's important to note that the power required for refrigeration is not just the power needed to freeze the water, but also the power needed to maintain the cold temperature. This is why the power required may seem low. In reality, it would require more power to continuously freeze water at a rapid rate than just to freeze it once.

Overall, your calculations are a good start, but make sure to use the correct units and consider all factors when determining the power required for refrigeration.