Calculate the amount of heat energy in kj required to convert 45.0 g of ice at -15.5'c to steam at 124.0°c. (cwater 118 jig'c, gee 2.03 jig c, g team jig c, molar heat of fusion of ice 6.01 * 10 j/mol; molar heat of vaporization of liquid water 4.07 * 10*j/mol 202 short answer toolbar navigation b i v s e 1 e a a this question will be sent to your instructor for grading
Answers: 1
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]](/tpl/images/0213/6323/e4ef0.png)
= enthalpy change or heat required = ?
= specific heat of solid water = 
= specific heat of liquid water = 
= specific heat of liquid water = 
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole![\Delta H=[45g\times 4.18J/gK\times (0-(-15.5))^oC]+2.5mole\times 6010J/mole+[45g\times 2.09J/gK\times (100-0)^oC]+2.5mole\times 40670J/mole+[45g\times 1.84J/gK\times (124-100)^oC]](/tpl/images/0213/6323/555ef.png)
(1 KJ = 1000 J)
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Calculate the amount of heat energy in kj required to convert 45.0 g of ice at -15.5'c to steam at 1...
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