can someone please help me figure out the followingproblem? 1.The heat of fusion

Question

can someone please help me figure out the followingproblem? 1.The heat of fusion of water is 80 cal/g, the heat ofvaporization of water is 540 cal/g, the specifc heat of ice is 0.48cal/deg/g, the specific heat of steam is 0.48 cal/deg/g and thespecific heat of liquid water is 1.0 cal/deg/g. How much heat wouldbe needed to convert 10.83 g of ice at -43 °C to steam at 136°C?

thank you!

can someone please help me figure out the followingproblem? 1.The heat of fusion of water is 80 cal/g, the heat ofvaporization of water is 540 cal/g, the specifc heat of ice is 0.48cal/deg/g, the specific heat of steam is 0.48 cal/deg/g and thespecific heat of liquid water is 1.0 cal/deg/g. How much heat wouldbe needed to convert 10.83 g of ice at -43 °C to steam at 136°C?

thank you!

Explanation / Answer

Ice at -43 C to ice at 0 C Heat required = mass * specific heat of ice * temperaturedifference                      = 10.83 g * 0.48 cal/C/g * (0 - (-43))                      = 223.53 cal Ice at 0 to water at 0 C Heat = mass * heat of fusion          = 10.83g * 80 cal/g          =866.4 cal Water at 0 to water at 100 C    Heat = 10.83 g * 1.0 cal/g.C * (100 -0)            =1083 cal Water at 100 C to steam at 100 C Heat = 10.83 g * 540 cal/g          =5848.2 cal Steam at 100 C to steam at 136 C Heat = 10.83g * 0.48 cal/ C.g * (136 - 100)          =187.14 cal To get the total heat, just add up the individual heats. To get the total heat, just add up the individual heats.

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