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

3, 360 J k g^{- 1}

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33, 600 J k g^{- 1}

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33, 60, 000 J k g^{- 1}

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3, 36, 000 J k g^{- 1}

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Solution

The correct option is C $3, 36, 000 J k g^{- 1}$
heat required to raise temperature is Q
$Q = 777000 J = 1 \times 2100 \times 10 + 1 \times L + 1 \times 4200 \times 100$
$L = 336000 J k g^{- 1}$

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Similar questions

$Calculate the amount of heat required to convert 1.00 kg of ice at -10°C$
$at normal pressure to steam at 100°C.$

$The specific heat capacity of ice = 2100 J k g^{- 1} K^{- 1}$

$Latent heat of fusion of ice = 3.36 \times 10^{5} J k g^{- 1}$

$The specific heat capacity of water = 4200 J k g^{- 1} K^{- 1}$

$Latent heat of vaporization of water = 2.25 \times 10^{6} J k g - 1$

A refrigerator converts 100 g of water at $20^{o} C$ to ice at $- 10^{o} C$ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ , specific latent heat of ice is 336 J $g^{-} 1$ and the specific heat capacity of ice is 2.1 J $g^{- 1} K^{- 1}$ .

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}

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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The amount of heat energy required to convert 1 kg of ice at −10∘C to water at 100∘C is 7,77,000 J. Calculate the specific latent heat of ice. Specific heat capacity of ice 2100Jkg−1K−1, specific heat capacity of water = 4200Jkg−1K−1.

The correct option is C 3,36,000Jkg−1heat required to raise temperature is QQ=777000J=1×2100×10+1×L+1×4200×100L=336000Jkg−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 = $4200 J k g^{- 1} K^{- 1}$ .

The correct option is C $3, 36, 000 J k g^{- 1}$
heat required to raise temperature is Q
$Q = 777000 J = 1 \times 2100 \times 10 + 1 \times L + 1 \times 4200 \times 100$
$L = 336000 J k g^{- 1}$