48. The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram

Question

48. The heat of fusion of ice is 333 J/g (meaning 333 J is absorbed when 1 gram of ice melts). The heat of vaporization of liquid water at 100 °C is 2257 1/g. Calculate the change in enthalpy, AH, in kJ/mol, for these two processes. H20(s)--> H2O(l): AH = ? a. H20(I)--> H2O(g): AH = ? b. b. So how are these two numbers (the 333 and 2257) related to the heat of fusion and heat of vaporization given at the beginning of the handout? Using the values you just calculated, determine the number of grams of ice that can be melted by 0.800 kJ of heat. c.

Explanation / Answer

H2O(s) melts to becomes H2O(l) by absorbing the latent heat of fusion. Hence deltaH= heat of fusion= 333 J/gm

H2O(l) becomes H2O(v) by absorbing heat of vaproization. Hecne deltaH= 2257 J/gm

since heat is absorbed, deltaH is +ve.

2. 333 J/gm is heat of fusion and 2257 J/gm is heat of vaporization.

3. during melting 333 J/gm is required or .333 Kj/gm is required.

0.8 Kj of heat melts 0.8/0.33 gm of ice =2.42 gm of ice.

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