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Standard XII

Physics

Specific Heat Capacity

the specific ...

Question

the specific heat capacity of liquid.

0.3 \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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3 \times 10^{3} J k g^{- 1} K^{- 1}

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

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Solution

The correct option is B $3 \times 10^{3} J k g^{- 1} K^{- 1}$
Heat supplied by heater is $600 \times 100 = 60000 J$
As $H e a t$ $E n e r g y = m s (T_{2} - T_{1})$
So specific heat capacity is $\frac{60000}{4 \times (15 - 10)} = \frac{60000}{20} = 3 \times 10^{3} J k g^{- 1} K^{- 1}$

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Similar questions

250 g

of water at

30^{o} 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^{o} 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}

2 k g

of ice melts when water at

100^{o} C

is poured in a hole drilled in a block of ice. What mass of water was used?

Given: specific heat capacity of water

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

specific latent heat of ice

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

Q. A copper calorimeter of mass

100 g m

contains

200 g m

of a mixture of ice and water. Steam at

100^{\circ} C

under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to

50^{\circ} C

. If the mass of the calorimeter and its contents is now

330 g m

, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper

= 0.42 \times 10^{3} J k g^{- 1} K^{- 1}

,
Specfic heat capacity of water

= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}

Specific heat of fusion of ice

= 3.36 \times 10^{5} J k g^{- 1}

Latent heat of condensation of steam

= 22.5 \times 10^{5} J k g^{- 1}

Q. If

10 g

of ice is added to

40 g

of water at

15^{o} C

, then the temperature of the mixture is (specific heat of water

= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}

,Latent heat of fusion of ice

= 3.36 \times 10^{5} J k g^{- 1}

)

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the specific heat capacity of liquid.

The correct option is B 3×103Jkg−1K−1Heat supplied by heater is 600×100=60000 JAs Heat Energy=ms(T2−T1)So specific heat capacity is 600004×(15−10)=6000020=3×103Jkg−1K−1

The correct option is B $3 \times 10^{3} J k g^{- 1} K^{- 1}$
Heat supplied by heater is $600 \times 100 = 60000 J$
As $H e a t$ $E n e r g y = m s (T_{2} - T_{1})$
So specific heat capacity is $\frac{60000}{4 \times (15 - 10)} = \frac{60000}{20} = 3 \times 10^{3} J k g^{- 1} K^{- 1}$