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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.


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Respuesta :

kkelhi

Answer:

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

Explanation: