The specific heat of water Sw-4200Jkg-1K-1 and the latent heat of ice Lf-3-4 - 105 Jkg-1K-1- 100 grams of ice at 0- is placed in 200 g of water at 0- The amount of ice that will melt as the temperature of water reaches 0- is close to in g

Here, ice melts due to energy released by water when it drops from 25^∘ to nbsp;0^∘ ∴ m_wS_wΔθ =m_iceL_ice nbsp; ⇒ m_ice=m_wS_wΔθL_ice =0.2× 4200× 253.4× ...

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Standard XII

Chemistry

Properties of Water

The specific ...

Question

The specific heat of water $S_{w} = 4200$ ${Jkg}^{- 1} K^{- 1}$ and the latent heat of ice $L_{f} = 3.4 \times 10^{5} {Jkg}^{- 1} K^{- 1} .$ $100$ grams of ice at $0^{\circ}$ is placed in $200 g$ of water at $0^{\circ} .$ The amount of ice that will melt as the temperature of water reaches $0^{\circ}$ is close to (in $g$ )

61.7

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63.8

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69.3

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64.6

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Solution

The correct option is A $61.7$
Here, ice melts due to energy released by water when it drops from $25^{\circ}$ to $0^{\circ}$

$∴ m_{w} S_{w} Δ θ = m_{i c e} L_{i c e}$

$\Rightarrow m_{i c e} = \frac{m_{w} S_{w} Δ θ}{L_{i c e}}$

$= \frac{0.2 \times 4200 \times 25}{3.4 \times 10^{5}} = 0.0617 kg = 61.7 g$

 Hence, $(A)$ is the correct answer.

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Similar questions

10

g of ice at

0^{\circ} C

absorbs

5460

J of heat energy to melt and change to water at

50^{\circ} C

. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is

4200 J k g^{- 1} K^{- 1}

The amount of heat energy required to convert 1 kg of ice at $- 10^{\circ} C$ to water at $100^{\circ} C$ is 7,77,000 J. Calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J $k g^{- 1} K^{- 1}$ , specific heat capacity of water=4200 J k $g^{- 1} K^{- 1}$ .

Q. A piece of ice of mass

40 g

0^{o} C

is added to

200 g

of water at

50^{\circ} C

. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

and specific latent heat of fusion of ice =

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

2 k g

of ice melts when water at

100^{\circ} C

is poured in a hole drilled in a block of ice. What mass of water was used?

Given: specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

specific latent heat of ice =

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

10 g of ice at 0 $^{\circ} C$ absorbs 5460 J of heat energy to melt and change t water at 50 $^{\circ} C$ . Calculate the specific latent heat of fusion of ice. (Specific heat capacity of water is 4200 J $K g^{- 1} K^{- 1}$ .

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The specific heat of water Sw=4200Jkg−1K−1 and the latent heat of ice Lf=3.4×105 Jkg−1K−1. 100 grams of ice at 0∘ is placed in 200 g of water at 0∘. The amount of ice that will melt as the temperature of water reaches 0∘ is close to (in g)

The correct option is A 61.7Here, ice melts due to energy released by water when it drops from 25∘ to 0∘ ∴mwSwΔθ=miceLice ⇒mice=mwSwΔθLice =0.2×4200×253.4×105=0.0617 kg=61.7 g  Hence, (A) is the correct answer.