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Standard XII

Physics

Specific Heat Capacity

An electric h...

Question

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 $^{o}$ C to 15.0 $^{o}$ C in 100 s. Calculate the specific heat capacity of liquid:

3 \times 10^{3} J k g^{- 1} K^{- 1}

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12 \times 10^{3} J k g^{- 1} K^{- 1}

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1.2 \times 10^{3} J k g^{- 1} K^{- 1}

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None of the above

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Solution

The correct option is A $3 \times 10^{3} J k g^{- 1} K^{- 1}$
Heat supplied by water is, $Q = 600 \times 100$
Also, $Q = m c Δ T$
Heat capacity, $c = Q / m Δ T = 3 \times 10^{3} J k g^{- 1} K^{- 1}$

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Similar questions

Q. An electric heater of power

600

W raises the temperature of

4.0

kg of a liquid from

{10.0}^{0} C

{15.0}^{0} C

100

s. Calculate: (i) The heat capacity of

4.0

kg of liquid, and (ii) the specific heat capacity of liquid.

An electric heater of power 600 W rasies the temperature of 4.0 kg of a liquid from ${10.0}^{\circ} C t o {15.0}^{\circ} C$ in 100 s.Calculate:(i) the heat capacity of 4.0 kg of liquid,and(ii) the specific heat capacity of liquid.

250 g

of water at

30^{\circ} C

is contained in a copper vessel of mass

50 g

. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to

5^{\circ} C

Given : specific latent heat of fusion of ice =

336 \times 10^{3} J k g^{- 1}

, specific heat capacity of copper =

400 J k g^{- 1} K^{- 1}

, specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

Q. Calculate the heat required to convert 

3 k g

of ice

a t - 12^{\circ} C

kept in a calorimeter to steam at

100^{\circ}

at atmospheric pressure. Given : specific heat of ice

= 2100 \times 10^{5} J {k g}^{- 1} K^{- 1}

, specific heat of water

= 4186 \times 10^{3} J {k g}^{- 1} K^{- 1}

, latent heat of fusion of

c e = 3.35 \times 10^{5} J {k g}^{- 1}

and latent heat of steam

= 2.256 \times 10^{6} J {k g}^{- 1}

Q. A piece of ice of mass

40 g

0^{o} C

is added to

200 g

of water at

50^{\circ} C

. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =

4200 J k g^{- 1} K^{- 1}

and specific latent heat of fusion of ice =

336 \times 10^{3} J k g^{- 1}

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An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0oC to 15.0oC in 100 s. Calculate the specific heat capacity of liquid:

The correct option is A 3×103 Jkg−1K−1Heat supplied by water is, Q=600×100Also, Q=mcΔTHeat capacity, c=Q/mΔT=3×103Jkg−1K−1

An electric heater of power 600 W raises the temperature of 4.0 kg of a liquid from 10.0 $^{o}$ C to 15.0 $^{o}$ C in 100 s. Calculate the specific heat capacity of liquid:

The correct option is A $3 \times 10^{3} J k g^{- 1} K^{- 1}$
Heat supplied by water is, $Q = 600 \times 100$
Also, $Q = m c Δ T$
Heat capacity, $c = Q / m Δ T = 3 \times 10^{3} J k g^{- 1} K^{- 1}$