Answer to Question #170062 in General Chemistry for Dalri

Question

Compare the number of calories absorbed when 578 g of ice at 0°C is changed to liquid water at 38°C with the number of calories absorbed when 578 g of liquid water is warmed from 0°C to 38°C.

Amount of heat in cal..................?

With steps please

Answer 1

Q170062

Compare the number of calories absorbed when 578 g of ice at 0°C is changed to liquid water at 38°C with the number of calories absorbed when 578 g of liquid water is warmed from 0°C to 38°C.

Solution:

Let us find the heat absorbed in both case.

Case 1 : 578g of Ice at 0°C is changed to liquid water at 38°C

Here, first ice will melt and convert to liquid water at 0°C and then water will warm up till 38°C.

Case 2 : 578 g of liquid water is warmed from  0°C  to final temperature 38°C.

here, we are just increasing the temperature of water which was initially at  0°C  to final temperature 38°C.

More heat energy will be absorbed in the Case 1, since some energy is used up in melting the ice.

Case 1 : 578g of Ice at 0°C is changed to liquid water at 38°C.

Enthalpy of fusion of water at  0°C = ΔH_fusion = 334 J/g.

Specific heat capacity of water, s = 4.184 J/g °C.

Ice (0°C ) ==> Water (  0°C ) ==> Water ( 38°C)

Heat absorbed in melting of 578 g of ice = m * ΔH_fusion = 578 g * 334J/g

= 193052 Joules.

Heat absorbed in warming water from  0°C to 38°C , Q = m * s * ΔT

ΔT = final temperature - initial temperature

ΔT = 38°C - 0°C = 38°C .

Q = 578g * 4.184 J/g °C * 38°C = 91879 Joules .

Total heat absorbed in Case 1 = 193052J + 91879 J = 284950 Joules.

which in calories $= 284950 \space Joules * \frac{1\space calorie }{4.184\space Joules } = 68105 \space calories.$

Case 2 : 578 g of liquid water is warmed from  0°C  to final temperature 38°C.

Heat absorbed in warming water from  0°C to 38°C , Q = m * s * ΔT

ΔT = final temperature - initial temperature

ΔT = 38°C - 0°C = 38°C .

Q = 578g * 4.184 J/g °C * 38°C = 91879 Joules .

which in calories $= 91879 \space Joules * \frac{1\space calorie }{4.184\space Joules } = 21960\space calories.$

mass of ice/water given to us is in 3 significant figure, so our final answer must also be in 3 significant figure.

Answer :

Heat absorbed in converting 578 g of ice at 0°C to liquid water at 38°C = 68100 calories.

Heat absorbed in warming 578 g of liquid water from 0°C to 38° C = 22000 calories.

By comparing both of them we see that, heat absorbed in converting 578 g of ice to liquid water at 38 C, is greater by 46100 calories than the heat absorbed in warming 578 g of water from

from 0°C to 38°C.

Thank you,