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 3.3x10^4 J Kg^-1.

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

A. 0˚CB. -0.36˚CC. -3.0˚CD. -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.

To solve this problem, we need to understand how energy transfer works when we put a bottle of water in a freezer. Let’s break it down step by step, so it’s easy to follow.

Understanding the Scenario

We have a plastic bottle with 1 liter of water at 20˚C, and we place it in a freezer that is at -18˚C. The bottle allows energy to escape at a rate of 800 joules per minute for two hours. Our goal is to find out the final temperature of the water after this time.

Calculating Total Energy Loss

First, we need to find out how much energy the water loses in two hours. Since the bottle loses energy at 800 joules per minute, we can calculate the total energy lost over two hours (which is 120 minutes).

• Energy lost per minute = 800 J
• Total time = 120 minutes
• Total energy lost = 800 J/min × 120 min = 96,000 J

Understanding Water's Properties

Next, we need to consider the properties of water. The specific heat capacity of water is 4200 J/kg·K, which means it takes 4200 joules to raise or lower the temperature of 1 kilogram of water by 1 degree Celsius. Since 1 liter of water has a mass of 1 kg, we can use this value directly.

Calculating Temperature Change

Now, we can find out how much the temperature of the water changes due to the energy loss. We can use the formula:

Q = mcΔT

Where:

• Q = energy lost (96,000 J)
• m = mass of water (1 kg)
• c = specific heat capacity (4200 J/kg·K)
• ΔT = change in temperature (in °C)

Rearranging the formula to find ΔT gives us:

ΔT = Q / (mc)

Substituting the values:

ΔT = 96,000 J / (1 kg × 4200 J/kg·K) = 22.86 K

Finding the Final Temperature

The temperature change we calculated is approximately 22.86°C. Since the water started at 20°C, we need to subtract this temperature change from the initial temperature:

Final Temperature = Initial Temperature - ΔT

Final Temperature = 20°C - 22.86°C = -2.86°C

Choosing the Closest Answer

Now, looking at the multiple-choice options provided:

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

The final temperature we calculated (-2.86°C) is closest to option C, which is -3.0°C. Therefore, the correct answer is:

C. -3.0˚C

Summary

In summary, by calculating the total energy lost and using the specific heat capacity of water, we determined that the water's temperature dropped to approximately -2.86°C after two hours in the freezer, which corresponds to the answer -3.0°C from the options given.

Approved