Calculate the amount of heat required to convert $500 g$ of ice into water without change of temperature ( $L_{i c e} = 3.34 \times 10^{5} J {k g}^{- 1}$ ).

1.67 \times 10^{5} J

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1.50 \times 10^{5} J

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1.26 \times 10^{5} J

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1 \times 10^{5} J

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Solution

The correct option is A $1.67 \times 10^{5} J$
Given , $L_{i c e} = 3.34 \times 10^{5} J / k g, m = 500 g = 0.5 k g$
From the definition of latent heat ,
$Q = m L$
or $Q = 0.5 \times 3.34 \times 10^{5} = 1.67 \times 10^{5} J$

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Similar questions

Q. Calculate the quantity of heat required to convert 1.5 kg of ice at

100^{\circ} C

to water at

15^{\circ} C

. (

L_{i c e} = 3.34 \times 10^{5} J {k g}^{- 1}

C_{w a t e r} = 4180 J {k g}^{- 1}^{\circ} C^{- 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

Q. What is meant by saying that the latent heat of fusion of ice is 3.34 × 10⁵ J/kg ?

The latent heat of fusion of ice is:

(a) 33.4 $\times 10^{5} J / K g$ (b) 22.5 $\times 10^{5} J / K g$
(c) 33.4 $\times 10^{4} J / K g$ (d) 2.25 $\times 10^{4} J / K g$

Why is the latent heat of fusion of ice i.e 3.34×10⁵J/kg less than the latent heat of vapourisation i.e 22.5×10⁵J/kg? Particles of solids are more tightly packed than liquids so literally the should require more energy for separation than liquids.

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Calculate the amount of heat required to convert 500g of ice into water without change of temperature (Lice=3.34×105Jkg−1).

The correct option is A 1.67×105JGiven , Lice=3.34×105J/kg,m=500g=0.5kgFrom the definition of latent heat , Q=mLor Q=0.5×3.34×105=1.67×105J

Calculate the amount of heat required to convert $500 g$ of ice into water without change of temperature ( $L_{i c e} = 3.34 \times 10^{5} J {k g}^{- 1}$ ).

The correct option is A $1.67 \times 10^{5} J$
Given , $L_{i c e} = 3.34 \times 10^{5} J / k g, m = 500 g = 0.5 k g$
From the definition of latent heat ,
$Q = m L$
or $Q = 0.5 \times 3.34 \times 10^{5} = 1.67 \times 10^{5} J$