Question

After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 5.50 kg of ice and the temperature of the ground was 43.0 °C, calculate the energy that is required to melt all the ice at 0 °C. The heat of fusion for water is 80.0 cal/g.

Answer 1

Answer:

$Q = 4.40 \times 10^5 Cal$

Explanation:

Here we know that initial temperature of ice is given as

$T = 0^o C$

now the latent heat of ice is given as

$L = 80 Cal/g$

now we also know that the mass of ice is

$m = 5.50 kg$

so here we know that heat required to change the phase of the ice is given as

$Q = mL$

$Q = (5.50 \times 10^3)(80)$

$Q = 4.40 \times 10^5 Cal$