Calculate the quantity of heat required to convet 1.5 kg of ice at $0^{o} C$ to water at $15^{o} C . (L_{i c e} = 3.34 \times 10^{5} J k g^{- 1}, C_{w a t e r} = 4180 J k g^{- 1}^{o} C^{- 1})$

5.85 \times 10^{5} J

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5.95 \times 10^{5} J

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3.95 \times 10^{5} J

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4.95 \times 10^{5} J

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Solution

The correct option is B $5.95 \times 10^{5} J$
$m = 1.5 k g$
$L = 3.34 \times 10^{5} J / k g$
$C = 4180 J / k g^{o} C$
$t_{1} = 0^{o} C, t_{2} = 15^{o} C$
Quantity of heat required, $Q = ?$
From relation heat required to current ice into water at $0^{o} C$
$Q_{1} = m L = 1.5 \times 3.34 \times 10^{5}$
$= 5.01 \times 10^{5} J$
Heat required is rise the temperature of water from $0^{o} C$ to $15^{o} C$
$Q_{2} = m C Δ t$
$= 1.5 \times 4180 \times (15 - 0) J$
$= 1.5 \times 4180 \times 15 J = 94050 J$
$= 0.94 \times 10^{5} J$
$∴$ Total heat required
$Q = Q_{1} + Q_{2}$
$= 5.01 \times 10^{5} + 0.94 \times 10^{5} J$
$= 5.95 \times 10^{5} J$ .

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Similar questions

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.

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

The latent heat of vaporisation of water is:
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(c) $2.25 \times 10^{4} J / K g$ (d) $33.4 \times 10^{5} J / K g$

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Calculate the quantity of heat required to convet 1.5 kg of ice at 0oC to water at 15oC.(Lice=3.34×105Jkg−1,Cwater=4180Jkg−1oC−1)

Calculate the quantity of heat required to convet 1.5 kg of ice at $0^{o} C$ to water at $15^{o} C . (L_{i c e} = 3.34 \times 10^{5} J k g^{- 1}, C_{w a t e r} = 4180 J k g^{- 1}^{o} C^{- 1})$