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=4186 J kg-1 K-1)

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 is 5°C the heat of fusion of ice is (swater=4186 J kg-1 K-1) (A) 1.42 × 105 J kg-1 (B) 2.42 × 105 J kg-1 (C) 3.42 × 105 J kg-1 (D) 4.42 × 105 J kg-1

Tardigrade Admin · Accepted Answer

Heat lost by water =mwsw(Ti-Tf) =2 × 4186 ×(70-5) =544180 J Heat required to melt ice =miLf=1.5 × Lf Heat required to rise temperature of ice =misw (Tf-T0)=1.5 × (4186) ×(5-0°) =31395 J By the principle of calorimetry Heat lost = heat gained 544180=1.5 Lf+31395 ∴ Lf=(512785/1.5)=341856.67 =3.42 × 105 J kg-1

Q. When $1.5 k g$ of ice at $0^{\circ} C$ mixed with $2 k g$ of water at $7 0^{\circ} C$ in a container, the resulting temperature is $5^{\circ} C$ the heat of fusion of ice is $(s_{w a t er} = 4186 J k g^{- 1} K^{- 1})$

$1.42 \times 1 0^{5} J k g^{- 1}$

$2.42 \times 1 0^{5} J k g^{- 1}$

$3.42 \times 1 0^{5} J k g^{- 1}$

$4.42 \times 1 0^{5} J k g^{- 1}$

Q. When 1.5kg of ice at 0∘C mixed with 2kg 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)

1.42×105Jkg−1

2.42×105Jkg−1

3.42×105Jkg−1

4.42×105Jkg−1

Q. When $1.5 k g$ of ice at $0^{\circ} C$ mixed with $2 k g$ of water at $7 0^{\circ} C$ in a container, the resulting temperature is $5^{\circ} C$ the heat of fusion of ice is $(s_{w a t er} = 4186 J k g^{- 1} K^{- 1})$

$1.42 \times 1 0^{5} J k g^{- 1}$

$2.42 \times 1 0^{5} J k g^{- 1}$

$3.42 \times 1 0^{5} J k g^{- 1}$

$4.42 \times 1 0^{5} J k g^{- 1}$