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Question: When 1.5 kg of ice at 0°C mixed with 2 kg of water at 70°C in a container, the resulting temperature...

When 1.5 kg of ice at 0°C mixed with 2 kg of water at 70°C in a container, the resulting temperature is 5°C the heat of fusion of ice is

(swater=4186Jkg−1K−1)\left( s_{water} = 4186Jkg^{- 1}K^{- 1} \right)

A

1.42×105Jkg−11.42 \times 10^{5}Jkg^{- 1}

B

2.42×105Jkg−12.42 \times 10^{5}Jkg^{- 1}

C

3.42×105Jkg−13.42 \times 10^{5}Jkg^{- 1}

D

4.42×105Jkg−14.42 \times 10^{5}Jkg^{- 1}

Answer

3.42×105Jkg−13.42 \times 10^{5}Jkg^{- 1}

Explanation

Solution

Heat lost by water = mwsw(Ti−Tf)m_{w}s_{w}(T_{i} - T_{f})

=2×4186×(70−5)=544180J= 2 \times 4186 \times (70 - 5) = 544180J

Heat required to melt ice =miLf=1.5×Lf= m_{i}L_{f} = 1.5 \times L_{f}

Heat required to rise temperature of ice

=misw(Tf−T0)= m_{i}s_{w}(T_{f} - T_{0})

=1.5×(4186)×(5−0∘)=31395J= 1.5 \times (4186) \times (5 - 0{^\circ}) = 31395J

By the principle of calorimetry

Heat lost = heat gained

544180=1.5Lf+31395544180 = 1.5L_{f} + 31395

∴Lf=5127851.5=341856.67=3.42×105Jkg−1\therefore L_{f} = \frac{512785}{1.5} = 341856.67 = 3.42 \times 10^{5}Jkg^{- 1}