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1st Law of Thermodynamics

An lnsulated ...

Question

An lnsulated vessel contains $0.4 k g$ of water at $0^{o} C$ . A piece of $0.1 k g$ ice at $- 15^{o} C$ is put into it and steam at $100^{\circ} C$ is bubbled into it until all ice is melted and finally the contents are liquid water at $40^{o} C$ . Assume that the vessel does not give or take any heat and there is no loss of matter and heat to the surroundings. Specific heat of ice is $2.2 \times 10^{3} J k g^{- 1} K^{- 1}$ , heat of fusion of water is $333 \times 10^{3} J k g^{- 1}$ and heat of vaporization of water is $2260 \times 10^{3} J k g^{- 1}$ . The amount of steam that was bubbled in to the water is about :

34.7 g

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236.0 g

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0.023 g

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48.0 g

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Solution

The correct option is D $48.0 g$
Mass of ice $m_{i} = 0.1 k g$
Mass of water $m_{w} = 0.4 k g$
Initial temperature of ice $T_{i} = - 15^{o} C$
Initial temperature of water $T_{1} = 0^{o} C$
Final temperature of water $T_{2} = 40^{o} C$

Total amount of heat absorbed $H_{a} = m_{i} S_{i} (T_{1} - T_{i}) + m_{i} L_{f} + (m_{i} + m_{w}) S_{w} (T_{2} - T_{1})$
$∴$ $H_{a} = 0.1 (2200) (15) + 0.1 (333000) + (0.5) (4200) (40) = 120600 J$

Heat released by the steam $H_{r} = m L_{v} + m S_{w} (100 - 40)$
$∴$ $H_{r} = m (2.260 \times 10^{3}) + m (4200) (60) = m (2512000) J$

Using $H_{a} = H_{r}$ we get $m \times 2.512 \times 10^{3} = 120600$ $⟹ m = 0.048 k g$
Hence mass of steam required is $48.0 g$

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Similar questions

Q. An Insulated vessel contains

0.4 k g

of water at

0^{o} C

. A piece of

0.1 k g

ice at

- 15^{o} C

is put into it and steam at

100^{o} C

is bubbled into it until all ice is melted and finally the contents are liquid water at

40^{o} C

. Assume that the vessel does not give or take any heat and there is no loss of matter and heat to the surroundings. Specific heat of ice is

2.2 \times 10^{3} J k g^{- 1} K^{- 1}

heat of fusion of water is

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heat of vaporization of water is

2260 \times 10^{3} J k g^{- 1}

. The amount of steam that was bubbled into the water is :

Q. A piece of ice of mass

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and specific latent heat of fusion of ice =

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Q. A copper calorimeter of mass

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336 \times 10^{3} J k g^{- 1}

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400 J k g^{- 1} K^{- 1}

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