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a piece of ice (heat capacity = 2100 JKg-10C-1 ) and latent heat = 3.36*105 JKg-1 of mass m gram is at -5 degree celsius at atm. it is given 420 J of heat so that the ice starts melting. finally when the mix. is in equilibrium , it is found that 1 gm ice has melted. assuming no heat change in the process, find the value of m.

kaajal , 10 Years ago
Grade 12
anser 1 Answers
Saurabh Kumar

Last Activity: 10 Years ago

Heat given = Heat required to melt 1 gm of ice at constant temp + Heat required to raise the temp of ‘m’ gm of ice.
420 J = (1/1000 Kg X 3.36x105) + (m/1000x2100x5) ..
Solve this equation and calculate the answer in Kg.

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