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Standard XII

Physics

Specific Heat Capacity

40 g of ice a...

Question

$40 g$ of ice at $0^{o} C$ is used to bring down the temperature of certain mass of water at $60^{o} C t o 10^{o} C$ . Find the mass of water used.
[Specific heat capacity of water $= 4200 J K g^{- 1} C^{- 1}$
[Specific latent heat of fusion of ice $= 336 \times 10^{3} J k g^{- 1}$ ]

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Solution

Let, mass of water used $= m$
Since, Heat gained $=$ Heat lost
$m_{1} c Δ t_{1} = m L + m_{2} c Δ t_{2}$
$m \times 4.2 \times (60 - 10) = (40 \times 336) + (40 \times 4.2 \times (10 - 0))$
$m \times 4.2 \times 50 = (40 \times 336) + 1680$
Therefore, the mass of water used, $m = \frac{13440 + 1680}{42 \times 5} = \frac{15120}{210}$ $= 72 g$

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Similar questions

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}

250 g

of water at

30^{o} C

is contained in a copper vessel of mass

50 g

. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to

5^{o} C

. Given : specific latent heat of fusion of ice

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

, specific heat capacity of copper

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

, specific heat capacity of water

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

2 k g

of ice melts when water at

100^{o} 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}

250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C.

Specific latent heat of fusion of ice = 336 x 10³ J kg^-1

Specific heat capacity of copper vessel= 400 J kg^-1 °C^-1

Specific heat capacity of water 4200 J kg^-1 °C^-1

40 g of ice at 0 $^{\circ} C$ is used to bring down the temperature of a certain mass of water from 60 $^{0} C$ to 10 $^{0} C$ . Find the mass of water used. Given, $S_{w} = 4200 J k g^{- 1}^{\circ} C^{- 1} a n d L_{f} = 336 \times 10^{3} j k g^{- 1}$

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40g of ice at 0oC is used to bring down the temperature of certain mass of water at 60oCto10oC. Find the mass of water used.[Specific heat capacity of water =4200 JKg−1C−1[Specific latent heat of fusion of ice =336×103Jkg−1]

Let, mass of water used =m Since, Heat gained = Heat lostm1cΔt1=mL+m2cΔt2m×4.2×(60−10)=(40×336)+(40×4.2×(10−0))m×4.2×50=(40×336)+1680Therefore, the mass of water used, m=13440+168042×5=15120210=72g

$40 g$ of ice at $0^{o} C$ is used to bring down the temperature of certain mass of water at $60^{o} C t o 10^{o} C$ . Find the mass of water used.
[Specific heat capacity of water $= 4200 J K g^{- 1} C^{- 1}$
[Specific latent heat of fusion of ice $= 336 \times 10^{3} J k g^{- 1}$ ]