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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

Zeba ansari , 7 Years ago

Grade 11

3 Answers

Arun

Last Activity: 7 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

Last Activity: 4 Years 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

Last Activity: 3 Years 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!