Calculate the quantity of heat required to convert 10 kg of ice at 0- to water at 50- C would be- Specific heat of water is 4200 J-kg - C-

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Standard X

Physics

Calorimeter

Calculate the...

Question

Calculate the quantity of heat required to convert 10 kg of ice at $0^{\circ}$ to water at $50^{\circ} C$ would be. Specific heat of water is 4200 J/kg $^{\circ} C$ .

4506 kJ

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3360 kJ

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5460 J

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5460 kJ

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Solution

The correct option is C 5460 kJ
we know that,
$W = m L + m c θ$
W= 10(336+ 4.2(50))= 5460 kJ
so quantity of heat required is 5460 J

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Similar questions

Q. Calculate the quantity of heat required to convert

10 k g

of ice at

0^{o} C

to water at

50^{o} C

(Specific heat of water is

4200 J / k g^{o} C

, Latent heat of fusion of ice is

336 \times 10^{3} J / k g

)

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 $g^{- 1} K^{- 1}$ .

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

10

g of ice at 0

^{o}

C absorbs

5460

J of heat energy to melt and change to water at

50

^{o}

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

4200

J kg

^{- 1}

^{- 1}

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Calculate the quantity of heat required to convert 10 kg of ice at 0∘ to water at 50∘C would be. Specific heat of water is 4200 J/kg ∘C.

The correct option is C 5460 kJwe know that,W=mL+mcθW= 10(336+ 4.2(50))= 5460 kJso quantity of heat required is 5460 J

Calculate the quantity of heat required to convert 10 kg of ice at $0^{\circ}$ to water at $50^{\circ} C$ would be. Specific heat of water is 4200 J/kg $^{\circ} C$ .

The correct option is C 5460 kJ
we know that,
$W = m L + m c θ$
W= 10(336+ 4.2(50))= 5460 kJ
so quantity of heat required is 5460 J