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.

help pls

Answer 1

Q = ΔH fusion * mass (g)

when we have:

ΔH fusion (the heat (or enthalpy) of fusion = 0.334 kJ/g

and mass of ice = 22.4 g

so by substitution, we can get the energy (Q) required to melt this mass of ice:

∴ Q = 0.334KJ/g * 22.4 g

= 7.48 KJ

∴ the energy required to melt 22.4 g of ice is = 7.48 KJ

Step-by-step explanation: