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Calculate the mass of ice needed to cool 150g of water contained in a calorimeter of mass 50g at 32 degree C such that the final temperature is 5 degree C. Specific heat capacity of calorimeter = 0.4 j/g degree C Specific heat capacity of water = 4.2 j/g degree C Latent heat capacity of ice = 330j/g
Asked by Sanyajbp | 04 Mar, 2019, 12:48: AM
answered-by-expert Expert Answer
Heat loss by (water+Calorimeter) = Heat gain by ice
 
Heat loss by (water+calorimeter) = mw Cpw ΔT + mC Cpc ΔT = mi ( L + Cpw δT )  ......................(1)
 
where,  mw = mass of water = 50 g
             Cpw  = Spcific heat of water = 4.2 J/( g °C )
             mC = mass of calorimeter = 50 g
             Cpc = Specific heat capacity of calorimeter = 0.4 J/( g °C )
             ΔT  = fall in temperature of water and calorimeter = 32-5 = 27°C
             mi = mass of ice in gram
             L = latenet heat capacity of ice = 330 J/g
             δT = rise in temperature = 5 °C
 
by substituting all the values in eqn.(1),  we get mass of ice ,    begin mathsize 12px style m subscript i equals space fraction numerator 150 cross times 4.2 cross times 27 space plus space 50 cross times 0.4 cross times 27 over denominator 330 cross times 4.2 cross times 5 end fraction space equals space 50 end style g
Answered by Thiyagarajan K | 04 Mar, 2019, 08:43: AM

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