A calorimeter of mass 60 g contains 180 g of water at 29oC. Find the final temperature of the mixture, when 37.2 g of ice at –10oC is added to it (specific heat capacity of water = 4200 J/kg K, latent heat of ice = 336x103 J/kg, specific heat capacity of ice = 2100 J/kg K, specific heat capaicity of the calorimeter is 0.42 Jg-1 oC-1).
Answers
Answered by
13
Heat gained=Heat lost
(mct)ice+(ml)=(mct)water+(mct)calorimeter
0.0372×2100×(T(-10))+0.0372×336×10^3=0.180×4200×(302-T)+60×420×(302-T)
Solve for T. I'll leave the calculations for you ^_^
(mct)ice+(ml)=(mct)water+(mct)calorimeter
0.0372×2100×(T(-10))+0.0372×336×10^3=0.180×4200×(302-T)+60×420×(302-T)
Solve for T. I'll leave the calculations for you ^_^
poojapraveen:
how did u get (302-t)
Yes,how did u get (302-t)....pls tell...
Similar questions