Marvin Sebourn

2019-02-11 03:21:38 UTC

We start with pure H2O.

10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.

40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.

Both ice and water are in separate, perfectly insulated containers with no void for air.

The ice and the water are then combined into one single, perfectly insulated container with no void for air.

Given:

The specific heat of water = 1 calorie per gram

The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.

Atmospheric pressure is normal, and constant at 14.7 psi.

When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?

Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

Thanks

Marvin Sebourn

***@aol.com

10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.

40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.

Both ice and water are in separate, perfectly insulated containers with no void for air.

The ice and the water are then combined into one single, perfectly insulated container with no void for air.

Given:

The specific heat of water = 1 calorie per gram

The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.

Atmospheric pressure is normal, and constant at 14.7 psi.

When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?

Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

Thanks

Marvin Sebourn

***@aol.com