A block of ice of mass 120   g at temperature 0 ° C is put in 300   g of water at 25 ° C . The x   g of ice melts as the temperature of the water reaches 0 ° C . The value of x is [Use: Specific heat capacity of water = 4200   J   kg - 1   K - 1 ,…

Question

A block of ice of mass 120 &#160; g at temperature 0 &#176; C is put in 300 &#160; g of water at 25 &#176; C . The x &#160; g of ice melts as the temperature of the water reaches 0 &#176; C . The value of x is [Use: Specific heat capacity of water = 4200 &#160; J &#160; kg - 1 &#160; K - 1 , Latent heat of ice = 3 . 5 &#215; 10 5 &#160; J &#160; kg - 1 ]

Aicharya · Accepted Answer

According to the principle of calorimetry, heat lost by hot body is equal to the heat gained by cold body.
The heat released by water
$∆ Q = m s ∆ T = 0.3 \times 4200 \times 25 = 31500 J$
let $m kg$ ice melts then,
$∆ Q = m L = m \times 3.5 \times 10^{5} = 31500 J$
$\Rightarrow m = \frac{31500 \times 10^{- 5}}{3.5} = 9000 \times 10^{- 5}$
$m = 0.09 kg = 90 g$
$\Rightarrow x = 90$

A block of ice of mass 120   g at temperature 0 ° C is put in 300   g of water at 25 ° C…

Solution

A block of ice of mass 120 &#160; g at temperature 0 &#176; C is put in 300 &#160; g of water at 25 &#176; C…

Solution

A block of ice of mass 120 g at temperature 0 ° C is put in 300 g of water at 25 ° C…