Question

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.

Answer 1

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

m = 32.34 pounds of ice.

Hope this helps