A plastic bottle containing a liter of water at 20˚C is placed inside a freezer, whose temperature is maintained at -18˚C.

The plastic bottle limits the transfer of energy to a mean value of 800 joule per minute for two hours and has negligible specific heat capacity.

For water, the specific heat capacity is 4200 J kg^-1 K^-1 and specific latent heat of fusion 3.3×10^4 J Kg^-1.

what is the temperature of water at the end of the two hours?

```
A. 0˚C
B. -0.36˚C
C. -3.0˚C
D. -18˚C
```

I couldn't do this question. It's multiple choice question and whatever I do, my answer is completely outside the 4 choices. Can anyone help? If possible, explain it as you would to a 5-year old, please.

## Best Answer

It seems like you have

notinvested enough effort and/or time into this question given the hundreds of better resources available on YouTube and Google. In my opinion, it would be better if you posted more specific questions in which you have shown exactly where you got stuck after a good amount of your own work and gone through more appropriate resources (YouTube is great at this). That being said, I hope you find the following helpful because (as I understand it), we don't do your homework questions for you here but we sometimes do guide you to the answer.Given the rate of energy transfer of $800J/min$, we can simplify this to $\frac{40}{3}J/sec$. If you take the estimate of 1 litre of water to be 1 kilogramme (this is not

exactlytrue, but it is very close), then you can calculate the temperature change of the $20$ degree Celsius water using the formula$Q=mc\Delta\theta$

and calculate the $Q$ produced (or in this case, taken away) by the fridge as

$\frac{40}{3} J/sec \times (2\times3600)$

because that rate of energy transfer is essentially a power equation ($P=\frac{\Delta E}{t}$). You would then equate that energy to the energy needed for lowering the temperature of the water with the aforementioned $Q=mc\Delta\theta$ where the $\Delta\theta$ is what you want to find out. You then realise that once you hit $0$ degrees Celsius, you have to use the other formula

$Q=mL_f$

because once liquid water hits $0$ degrees Celsius, it doesn't go lower but rather uses the energy to transition its phase into a solid one first (assuming this is at normal pressure). Note that here the Q here is the energy left over

afteryou have subtracted the energy used for lowering the temperature from the original energy calculated from the power equation.Ifthere is still leftover energy after the water has become ice, you then move to lower the temperature of the ice but this time using the specific heat capacity of ice. The question did not provide the specific heat capacity of ice (this should be a hint).The temperature of the freezer is a red herring here because we assume the freezer has more than enough "capacity" to cool the water at that same rate for two hours. To assume otherwise would be too hard for an A-Level question (they also say it goes on for two hours). There is also another

trickhere but that should go away with some reasoning of "What would happen if I had enough energy to go to $0$ but not enough tofreezethe whole bottle of water ?"