Calculate the total amount of heat in kilocalories needed to change 500 grams of

Question

Calculate the total amount of heat in kilocalories needed to change 500 grams of ice at -10 degrees C into 500g of steam at 120 degrees C
Specific heat of ice= 0.51 cal/g-degree Specific heat of liquid water = 1.00 cal/g-degree Specific heat of gas = 0.48 cal/g-degree Heat of fission of water = 80 cal/g Heat of vaporization of water = 540 cal/g Calculate the total amount of heat in kilocalories needed to change 500 grams of ice at -10 degrees C into 500g of steam at 120 degrees C
Specific heat of ice= 0.51 cal/g-degree Specific heat of liquid water = 1.00 cal/g-degree Specific heat of gas = 0.48 cal/g-degree Heat of fission of water = 80 cal/g Heat of vaporization of water = 540 cal/g
Specific heat of ice= 0.51 cal/g-degree Specific heat of liquid water = 1.00 cal/g-degree Specific heat of gas = 0.48 cal/g-degree Heat of fission of water = 80 cal/g Heat of vaporization of water = 540 cal/g

Explanation / Answer

The "cold water" in this case is going to do three things:

a) as a solid, warm up from -10 to zero
b) all 18 g will melt
c) as a liquid, the temperature goes to 100 Deg celcius

d) All liquid will convert to stream

e) As a stream it will reach the temperature 120 Deg cel

Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in Cal /(g oC)
T = change in temperature = Tfinal - Tinitial in oC

Q = 500 x 0.51 x 10 + 500 x 80/18 + 500 x 1 x 100 + 500 x 540 /18 + 500 x 0.48 x 20

Q = 2550 + 2222 + 50000 + 15000 + 4800

Q = 74602 Calories

Q = 74.602 Kilo Calories

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