Name: A Laboratory Thermometer
Uploaded: 2019-07-19
Duration: 4 min 53 s
Description: We need a device to measure temperatures more accurately in laboratories. In this video, we introduce you to a device that will help us measure laboratory te...

Question

A piece of ice of mass

60 g

is dropped into

140 g

of water at

50 ° C

. Calculate the final temperature of water when all the ice has melted. (Assume no heat is lost to the surrounding) Specific heat capacity of water

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

Specific latent heat of fusion of ice

= 336 J g^{- 1}

Accepted Answer

Given $m_{1} = 140 g = 140 \times 10^{- 3} kg, m_{2} = 60 g = 60 \times 10^{- 3} kg$

The temperature of the water $T_{1} = 50 ° C$ ,

The specific heat capacity of water $c = 4.2 J g^{- 1} K^{- 1}$

Specific latent heat of fusion of ice $L = 336 J g^{- 1}$

The specific latent heat of a substance is the amount of energy needed to change the state of one kilogram of the substance without changing the temperature whereas the latent heat of fusion is the amount of energy to freeze or melt the substance at its melting point.

Let the final temperature be $T ° C$

The heat lost by the water $=$ The heat absorbed by the ice.

$m_{1} c ∆ T = m_{2} L + m_{2} c ∆ T$

$140 \times 10^{- 3} \times 4.2 \times 10^{3} \times (50 - T) = 60 \times 10^{- 3} \times 336 \times 10^{3} + 60 \times 10^{- 3} \times 4.2 \times 10^{3} \times (T - 0)$

$588 \times (50 - T) = 20160 + 252 T$

$29400 - 588 T = 20160 + 252 T$

$T = \frac{9240}{840} = 11 ° C$ .

Hence, the final temperature of the water is $11 ° C$ .

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