You visited us 1 times! Enjoying our articles? Unlock Full Access!

Latent Heat of Fusion

In an experim...

Question

In an experiment, $17 g$ of ice is used to bring down the temperature of $40 g$ of water at $34^{0} C$ to its freezing temperature. The specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$ . Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

Open in App

Solution

Mass of ice $m_{1} = 17 g$
Mass of water $m_{2} = 40 g$ .
Change in temperature = 34 – 0 = 34K
Specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$ .
Assuming there is no loss of heat, heat energy gained by ice (latent heat of ice),
Q= heat energy released by water
Q = 40 × 34 × 4.2 = 5712 J.
Specific latent heat of ice =
$L = \frac{Q}{m} = \frac{5712}{17} = 336 J g^{- 1}$

Suggest Corrections

Similar questions

17

40

34

^{o}

4.2

^{- 1}

^{- 1}

100 g

- 10^{o} C

100^{o} C

= 2.1 J g^{- 1} K^{- 1}

= 4.2 J g^{- 1} K^{- 1}

= 336 J g^{- 1}

40 g

0^{o} C

60^{o} C t o 10^{o} C

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

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

250 g

30^{o} C

50 g

5^{o} C

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

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

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

100 g

20^{o} C

10^{o} C

73.5 m i n

4.2 J g^{- 1} K^{- 1}

336 J g^{- 1}

2.1 J g^{- 1} K^{- 1}

Join BYJU'S Learning Program