please will someone help me? The heat of fusion of water is 80 cal/g, the heat o

Question

please will someone help me?
The heat of fusion of water is 80 cal/g, the heat of vaporizationof water is 540 cal/g, and the specific heat of water is 1.0cal/deg/g. How many grams of ice at 0 ° could be converted tosteam at 100 °C by 9,746 calories of heat?

thank you please will someone help me?
The heat of fusion of water is 80 cal/g, the heat of vaporizationof water is 540 cal/g, and the specific heat of water is 1.0cal/deg/g. How many grams of ice at 0 ° could be converted tosteam at 100 °C by 9,746 calories of heat?

thank you

Explanation / Answer

So i will say we have 10 g. You can use what ever numberyou want. Will use a proportion at the end. Q = cmT Q = Lm melting Q = (80 cal/g)(10g) = 800 cal heating of liquid Q = (1.0 cal/deg/g)(10 g)(100 deg) = 1000 cal vaporization Q = (10 g)(540 cal/g) = 5400 cal for 10 g 7200 cal 10/7200 = x/9747 solve for x 13.538 g

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