Latent Heat Problem/ Phase Change .

[jdhutto]

Latent Heat Problem/ Phase Change...Please Help!

At 1:00pm you place an ice cube tray in the freezer. Each of the 12 1-gram cubes has a temperature of 20°C. At 1:10 the water temperature has dropped to 12°C. At what time will you have ice?

i have been messing around with this problem for like 20 minutes and I am missing something, I am using Q=mc delta T, and i know it takes 1 cal/g/degree C to go from 20 down to zero, and you have to add the latent heat from water to ice to that to get Q total, i just can't figure out how to work the time factor in...please help, thanks

 

[alphysicist]

Hi jdhutto,

At 1:00pm you place an ice cube tray in the freezer. Each of the 12 1-gram cubes has a temperature of 20°C. At 1:10 the water temperature has dropped to 12°C. At what time will you have ice?

i have been messing around with this problem for like 20 minutes and I am missing something, I am using Q=mc delta T, and i know it takes 1 cal/g/degree C to go from 20 down to zero, and you have to add the latent heat from water to ice to that to get Q total, i just can't figure out how to work the time factor in...please help, thanks

I believe that they want you to assume that the rate of heat flow out of the ice is constant. Do you see what to do now?

 

[baba89]

Based on the information provided, we can calculate the amount of energy required to freeze the water in the ice cube tray. First, we need to determine the specific heat capacity of water, which is 1 cal/g/°C. Next, we need to calculate the change in temperature from 20°C to 0°C, which is a decrease of 20°C. This gives us a total of 20 calories of energy needed to freeze 1 gram of water.

Since we have 12 cubes, we need to multiply 20 calories by 12, which gives us a total of 240 calories of energy needed to freeze all 12 cubes. Now, we need to factor in the latent heat of fusion, which is the amount of energy required to change water from a liquid to a solid at its freezing point. This value is 80 calories per gram of water.

Since we have 12 grams of water in each cube, we need to multiply 80 calories by 12, which gives us a total of 960 calories of energy needed for the phase change. Adding the 240 calories needed for the temperature change, we get a total of 1200 calories of energy needed to freeze all 12 cubes.

Now, we can use the formula Q = mct to calculate the time it takes for the water to freeze. We know the mass (12 grams), the specific heat capacity (1 cal/g/°C), and the change in temperature (20°C), so we can plug in these values to get:

1200 calories = (12 grams) x (1 cal/g/°C) x (t)

Solving for t, we get t = 100 seconds, which is approximately 1 minute and 40 seconds.

Therefore, the ice cubes should be frozen by 1:11pm. It is important to note that this calculation is based on ideal conditions and may vary in real life due to factors such as the efficiency of the freezer and the temperature of the surrounding environment. I hope this helps!