# Answer to Question #169059 in General Chemistry for SREDDA

Question #169059

If:

Specific heat of ice = 0.48 cal/g · °C

Specific heat of steam = 0.48 cal/g · °C

Heat of fusion of ice = 80. cal/g

Heat of vaporization = 540 cal/g

Specific heat of liquid water = 1.00 cal/ g · °C

At −13.0°C to water vapor at 112°C, What is the total heat in calories that is required to raise the temperature of 5.60 g of ice?

Total heat=...........cal

1
2021-03-05T03:31:58-0500

Split the temperature into three parts -13 to 0 , 0 to 100 and 100 to 112°C . Because there is phase change at 0 and at 100°C.

Heat = mC"\u0394T"

= 5.6 × 4.18 x 99 =2.3 KJ

In order to calculate total we need to add heat of vaporization and fusion

Heat of fusion = 80× 5.6x 4.18= 1.87 KJ

Heat of vaporization = 540x5.6x4.18= 12.64 KJ

Total heat = 12.64 + 1.87 + 2.3 = 16.81 KJ

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