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2 years ago

How many pounds of ice are required to absorb 4900 kJ of heat as the ice melts? The heat of fusion of water is 0.334 kJ/g.

Chemistry

1 answer:

Goshia [24]2 years ago

5 0

Answer:

m = 32.34 pounds of ice.

Explanation:

In this case we need to use the following expression of heat:

q = m * ΔHf (1)

Where:

q: heat absorbed in J or kJ

m: mass of the compound in g

ΔHf: heat of fusion of the water in kJ/g

We are asked to look for the mass of ice in pounds, so after we get the grams, we need to convert the grams to pounds, using the following conversion:

1 pound --------> 453.59 g (2)

So, we have the heat and heat of fusion, from (1) let's solve for the mass, and then, using (2) the conversion to pounds:

q = m * ΔHf

m = q / ΔHf

m = 4900 / 0.334 = 14,670.66 g of ice

Now, the conversion to pounds:

m = 14,670.66 g * 1 pound/453.59 g

<h2>m = 32.34 pounds of ice.</h2>

Hope this helps

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