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a calorimetry of mass 50g contains 200g of water at 30degree celcius . Calculate the final temperature of the mixture when 40g of ice at -10degree celcius is added to it.
Material of calorimeter or specific heat of material of calorimeter is not known.
Hence it is assumed heat loss of calorimeter is very less compared to heat loss of water.

Heat loss of water ,  mice × [ CP_ice × 10 + Lf + CPw ( 30 - T ) ] = mw × Cpw × ( 30 - T)

where mice = 40 g is mass of iice , CP_ice = 2040 J/KgK is specific heat of ice,
Lf = 335 kJ/kg is latent heat of fusion of ice , CPw = 4200 J/kgK is specific heat of water,
T is equilibrium temperature and mW = 200g is mass of water.

40 × 10-3 [ 2040 × 10 + 335 × 103 + 4200T) ] = 200 × 10-3 × 4200 × ( 30 - T )

we get T = 10.9 oC from above equation

Answered by Thiyagarajan K | 27 Feb, 2021, 11:54: PM

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