1 kg of ice at 0- C is mixed with 1 kg of steam at 100- C- What will be the composition of the system when thermal equilibrium is reached- Latent heatof fusion of ice -3-36 - 105 J kg -1 and latent heat of vaporization of water -2-26 - 106 J kg -1A- 2 kg iceB- 665 gm water- 1-335 kg steamC- 665 gm steam- 1-335 kg waterD- 2 kg steam

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Calorimetry

1 kg of ice a...

Question

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$

2kg steam

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2 kg ice

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665gm water, 1.335 kg steam

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665 gm steam, 1.335 kg water

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Solution

The correct option is D
665 gm steam, 1.335 kg water

As the mass of ice and steam is the same, let us start by assuming that all the ice converts to water at 0°C, then heats up to water at $100^{\circ} C$ , in the process converting some of the steam to water at $100^{\circ} C$ . Don't worry, if our assumption is incorrect, the mass of water will come out negative and we can then consider other possibilities!
So, Heat gained during ice at $0^{\circ} C$ $\to$ water at $100^{\circ} C$

$Q_{1}$ = $m L_{1} + m s Δ T$

$m$ = $1 k g$

$L_{1}$ $\to$ Latent heat of fusion
$Δ T$ = $100^{\circ} C$
$S \to$ specific heat of water
$\Rightarrow$ $Q_{1}$ = $L_{1} + 100 S$

Now, heat released by $x$ kg steam converted to water at $\to$ is

$Q_{2}$ = $x L_{2}$

Where, $L_{2}$ $\to$ latent heat of vaporization

But, $Q_{1}$ = $Q_{2}$

$L_{1} + 100 S$ = $x L_{2}$
$\Rightarrow$ $x$ = $\frac{L_{1} + 100 s}{L_{2}}$

$\Rightarrow$ $x$ = $0.335 k g$ steam converts to water

$∴$ Amount of steam left = $1 - 0.335 = 0. 665 k g$

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Similar questions

1 k g

of ice at

0^{\circ} C

is mixed with

1 k g

of steam at

100^{\circ} C

. What will be the composition of the system when thermal equilibrium is reached? Latent of fusion

3.36 \times 10 J / g

and latent heat of vaporization of water =

2.26 \times 10 J / g

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$

1 kg

of ice at

0^{\circ} C

is mixed with

1 kg

of steam at

100^{\circ} C

. Find the mass of steam when the mixture has reached thermal equilibrium.

[L_{f u s i o n, i c e} = 3.36 \times 10^{5} J/kg,

L_{v a p o r i z a t i o n, w a t e r} = 2.26 \times 10^{6} J/kg,

C_{w a t e r} = 4200 J/kg-K]

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

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1 kg of ice at 0∘C is mixed with 1 kg of steam at 100∘C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 105 J kg−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1

1 kg of ice at $0^{\circ} C$ is mixed with 1 kg of steam at $100^{\circ} C$ . What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = $3.36 \times 10^{5} J k g^{- 1}$ and latent heat of vaporization of water = $2.26 \times 10^{6} J k g^{- 1}$