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10g of ice at 0°c is mixed with 100 g of water of 50°c what is the resultant temperature of mixture

10g of ice at 0°c is mixed with 100 g of water of 50°c what is the resultant temperature of mixture

Grade:11

3 Answers

Arun
25763 Points
3 years ago
Dear student

Latent heat of fusion of ice = 336 J/g
specific heat of water = 4.186 J/g degreeC
First 10g ice melts and then its temperature increases by gaining heat from 100g water. let the resultant temperature is T.
Energy gained by 10g ice = energy lost by 10g water
10*(336) + 10*4.186*(T-0) = 100*4.186*(50-T)
3360 + 41.86 T = 20930 - 418.6T
(41.86+418.6)T = 20930 - 3360
460.46T = 17570
T = 38.15 degree celsius
 
Regards
Arun (askIITians forum expert)
Kushagra Madhukar
askIITians Faculty 629 Points
11 months ago
Dear student,
Please find the attached solution to your problem below.

Latent heat of fusion of ice = 336 J/g
specific heat of water = 4.186 J/g degreeC
First 10g ice melts and then its temperature increases by gaining heat from 100g water. let the resultant temperature is T.
Energy gained by 10g ice = energy lost by 10g water
10*(336) + 10*4.186*(T-0) = 100*4.186*(50-T)
3360 + 41.86 T = 20930 - 418.6T
(41.86+418.6)T = 20930 - 3360
460.46T = 17570
T = 38.15 degree celsius

Thanks for asking the question.
Regards,
Kushagra
AGBO OSONDU
15 Points
2 months ago
60 dkg water with 70 °C is mixed with 10 dkg ice with -10 °C. How much will the temperature of the mixed material, and how change the entropy of the system? Prove that this process is irreversible!

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