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Standard VII

Physics

Hot and Cold

A refrigerato...

Question

A refrigerator converts $100 g$ of water at $25^{\circ} C$ into ice at $- 10^{\circ} C$ in $1 h$ and $50 m i n$ . The quantity of heat (in calorie) removed per min is (specific heat of ice $s_{i c e} = 0.5 c a l / g^{\circ} C$ , specific heat of water $s_{w} = 1 c a l / g^{\circ} C$ latent heat of fusion $= 80 c a l / g$ )

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Solution

We know that: $Q = m L, Q = m s Δ θ$

Given: $m = 100 g, θ_{1} = 25^{\circ} C, θ_{2} = - 10^{\circ} C, t = 1 h 50$ min, $s_{i c e} = 0.5 c a l / g^{\circ} C, s_{w} = 1 c a l / g^{\circ} C, L = 80 c a l / g$

Heat released by water to reach ice will be

$Q = m s_{w} (25 - 0) + m L + m s_{i c e} (0 - (- 10))$

$Q = 100 \times 1 \times 25 + 100 \times 80 + 100 \times 0.5 \times 10$

$Q = 2500 + 8000 + 500 = 11000 c a l$

Heat removed per minute is

$\frac{11000}{110} = 100 c a l / m i n$

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Similar questions

Q. A piece of ice of mass

50 g

exists at a temperature of

- 20^{\circ} C

. Determine the total heat required to convert it completely to steam at

100^{\circ} C

. (Specific heat capacity of ice

= 0.5 c a l / g^{\circ} C

; specific latent heat of fusion for ice

= 80 c a l / g

and specific latent heat of vapourisation for water

= 540 c a l / g

)

Q. Work done in converting one gram of ice at

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into steam at

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is -
[Given, specific heat of ice

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

, specific heat of water

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

, latent heat of fusion of ice

L_{f} = 80 cal/g

, & latent heat of vaporization of steam

L_{v} = 540 cal/g

]

Q. Calculate the mass of steam at

100^{\circ} C

needed to just melt 320 g of ice at

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. The specific heat of ice

= 0.5 c a l g^{- 1}^{\circ} C^{- 1}

, latent heat of vaporization of steam

= 540 c a l g^{- 1},

specific heat of fusion of ice

= 80 c a l g^{- 1}

Q. The amount of heat required to convert

1 g

of ice at

- 10^{o}

C into steam at

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C is (specific heat of ice

= 0.5

cal/g/

^{o}

C, latent heat of ice

= 80

cal/g, latent heat of steam

= 540

cal/g)

1 kg

of ice at

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is mixed with

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. The final temperature of mixture is

Given specific heat of ice

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,
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= 1 {cal/g}^{\circ} C

,
Latent heat of fusion for ice

= 80 cal/g

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Hot and Cold

Standard VII Physics

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A refrigerator converts 100 g of water at 25∘C into ice at −10∘C in 1 h and 50 min. The quantity of heat (in calorie) removed per min is (specific heat of ice sice=0.5 cal/g∘C, specific heat of water sw=1 cal/g∘C latent heat of fusion =80 cal/g)