Years ago, a block of ice with a mass of about 22kg was used daily in a home icebox. The temperature of the ice was 0.0 degrees Celsius when delivered.

As it melted, how much heat did a block of ice that size absorb? The latent heat of fusion of water is 3.34 x 10 ^ 5 J/kg.
Answer in units of J.

heat lost by ice=MLf

=22*3.34*10^5 J
=7348000 J

since heat gained =heat lost
7348000 joules of heat energy is absorbed by the ice box.

Thank you!!!

mention not...

To find out how much heat a block of ice absorbed as it melted, you can calculate the latent heat of fusion multiplied by the mass of the ice.

The latent heat of fusion is given as 3.34 x 10^5 J/kg (joules per kilogram), and the mass of the ice is stated as 22 kg.

The formula to calculate the heat absorbed is:

Heat = latent heat of fusion x mass

Plugging in the values, we get:

Heat = (3.34 x 10^5 J/kg) x 22 kg

Now we can calculate the heat absorbed:

Heat = 7.348 x 10^6 J

Therefore, a block of ice with a mass of 22 kg absorbed approximately 7.348 x 10^6 joules of heat as it melted.