Answer to Question #170062 in General Chemistry for Dalri

Question #170062

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


1
Expert's answer
2021-03-10T06:16:22-0500

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 = ΔHfusion = 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 * ΔHfusion = 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,



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