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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.

Respuesta :

Answer:

[tex]Q = 4.40 \times 10^5 Cal[/tex]

Explanation:

Here we know that initial temperature of ice is given as

[tex]T = 0^o C[/tex]

now the latent heat of ice is given as

[tex]L = 80 Cal/g[/tex]

now we also know that the mass of ice is

[tex]m = 5.50 kg[/tex]

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

[tex]Q = mL[/tex]

[tex]Q = (5.50 \times 10^3)(80)[/tex]

[tex]Q = 4.40 \times 10^5 Cal[/tex]