104 g of water at 30°C is taken in a caloriemeter made of copper of mass 42g. When a certain mass of ice at 0°C is added fo it, the final steady temperature of the mixture after the ice has melted, was found to be 10°C. Find the mass of ice added.


 

Asked by rishi16.anvi | 11th Sep, 2019, 05:35: PM

Expert Answer:

Heat loss of (water+Copper calorimeter) = ( mw Cpw  + mC Cpc ) ΔT  ......................(1)
 
where,  mw = mass of water = 104× 10-3 kg
Cpw = Sp. Heat of water = 4200 J/ (kg oC)
mC = mass of calorimeter = 42×10-3 kg
Cpc = Sp.Heat of copper = 385 J/ (kg oC)
 
ΔT = temperature difference = 30 - 10 = 20 oC
 
Let m be the mass of ice added to water
 
Heat gain by ice = m×(L+Cpw × ΔT) = m ×(334×103+ 4200 × 10) ......................(2)
 
where L is latent heat of fusion of ice , L = 334 kJ/kg
 
By substituting all the values in eqn.(1) and  considering heat loss of water+Calorimeter is equal to heat gain by ice,
we solve for mass of ice m to get
 
m = [ ( 104 × 10-3 × 4200 + 42 ×10-3 × 385 )× 20 ] / (334×103 + 42000 )   = 0.024 kg = 24 g

Answered by Thiyagarajan K | 11th Sep, 2019, 07:48: PM

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