0.5kg of water at 10ºC is completely converted to ice at 0ºC by extracting (88000) of heat from it. If the specific heat capacity of water is 4200jkg1 1 Cº....

Answer Details

Heat involved (H) = 88,000J, Mass(M)= 0.5kg,  specific heat capacity of water(C) = 4200JkgºC,

Specific latent heat of fusion(L) = ?, Temperature change(Δθ) =  θ2 -  θ1 =  (10 - 0)° = 10°

H = MCΔθ + ML 

or

H = M(CΔθ + L) -->HM $\frac{H}{M}$ = CΔθ + L 

: L = HM $\frac{H}{M}$ -  CΔθ = 88,0000.5 $\frac{88,000}{0.5}$ - 4200 × $\times$ 10

L = 176,000 - 42000 = 134,000

L = 134,000 or 134kj/kg