Given the Following: heat of fusion of ice = 80.0 cal/g; specific heat of ice = 0.480 cal/g; and specific heat of water = 1.00 cal/g. Calculate the amount of heat required to melt 30.0 g of ice at 0 degrees celsius?

Question

Avery Adams · Accepted Answer

We wish to assess the enthalpy change for the physical reaction......

#H_2O(s) +DeltararrH_2O(l)#....... Note that BOTH the ice and water have a specified temperature of #0# #""^@C#, and thus we assess the heat involved in the phase change, and thus we only need the #"latent heat of fusion of ice"#, #80.0*cal*g^-1#. And for the given quantity, this #80*cal*g^-1xx30*g=2400*cal#; whatever a #"calorie"# is........... In the same vein, we would report the reverse reaction's enthalpy change. #H_2O(l) +DeltararrH_2O(s)#, #DeltaH^@=-2400*cal#. Do you agree?

Naomi Aronson · Answer

undefined To calculate the amount of heat required to melt 30.0 g of ice at 0 degrees Celsius, we need to consider two steps: first, we need to heat the ice from 0°C to its melting point, and then we need to melt the ice.

1. Heat required to raise the temperature of ice from 0°C to its melting point:

q1 = mass × specific heat of ice × temperature change
   = 30.0 g × 0.480 cal/g°C × (0°C - 0°C)
   = 0 cal

2. Heat required to melt the ice:

q2 = mass × heat of fusion of ice
   = 30.0 g × 80.0 cal/g
   = 2400 cal

Therefore, the total heat required is:
Total heat = q1 + q2
            = 0 cal + 2400 cal
            = 2400 cal

So, 2400 calories of heat are required to melt 30.0 g of ice at 0 degrees Celsius.