Question

A container of negligible heat capacity has in it 456 g of ice at - 25 . 0°C . Heat is supplied to the container at the rate of 1000 J / min . After how long will the ice start to melt , assuming all of the ice has the same temperature ? The specific heat of ice is 2090 J / ( kg K ) and the latent heat of fusion of water is 33 . 5 x 104J / kg .

Answer 1

The ice will start to melt after 25.4 mins.

Step-by-step explanation:

To melt the ice , heat should be supplied such that two processes take place-

i) Increasing the temperature of ice upto 0.0°C

ii) Phase change from ice to water at constant temperature of 0.0°C

Given that,

Initial temperature
`T_(1)` = -25.0°C

Final temperature
`T_(2)` = 0.0°C

mass of ice , m = 456 g = 0.456 kg

Specific heat of ice , s = 2090 J/(kg K)

Latent heat of fusion , L = 33.5 × 104 J/kg

For process i) , heat that should be supplied is -

`Q_(1) = ms\Delta T` =
`ms(T_(2)-T_(1))` = 0.456×2090×(0-(-25))

`Q_(1)` = 23826 J

For process ii) , heat that should be supplied is -

`Q_(2) = mL` = 0.456×33.5 × 104 = 1588.704 J

∴Total heat required is -

`Q_(1)` +
`Q_(2)` = 25414.704 J

Rate of heat supply (r) = 1000 J/min

∴ The required time (t) is given by -

`t = (25414.704)/(r)` =
`(25414.704)/(1000)` = 25.4 mins

∴ t = 25.4 mins