Question

A 20.0 kg block of ice is sitting in a cooler. The temperature of the ice is 0.0ºC when it is put into the cooler. As it melts, how much heat does the block of ice absorb? (Remember Hf of water is 3.34 X 105 J/kg)

Answer 1

To determine the amount of heat the block has absorbed, we simply multiply the mass of the block of ice to the heat of fusion of water which is given above. We calculate as follows:

Heat = 20.0 kg (3.34 x 10^5 J/kg)
Heat = 6680000 J

Hope this helped.

Answer 2

Answer : The amount of heat required is, (C) $6.68\times 10^6J$

Explanation :

Formula used :

$q=L\times m$

where,

q = heat required = ?

L = latent heat of fusion of ice = $3.34\times 10^5J/kg$

m = mass of ice = 20.0 kg

Now put all the given values in the above formula, we get:

$q=(3.34\times 10^5J/kg)\times (20.0kg)$

$q=6.68\times 10^6J$

Therefore, the amount of heat required is, $6.68\times 10^6J$