Question

0.5 kg of ice at 0°C is heated uniformly by an electric heater of power 2kW if all heat is absorbed by ice calculate the time intervals in seconds for (i) ice to completely melt to form water at 10°C (ii) water to attain a temperature of 100°C (iii) water to change to steam at 100°C given: Sp.latent heat of ice = 336000JKg^-1 Sp.latent heat of steam = 2260000JKg^-1 Sp.heat capacity of water = 4200JKg^-1K^-1​

Answer 1

Here the solutiona attached below.

However in second part It is not given to change water from 0°C or 10°C. So in brackets I have given the alternate form also.

Answer 2

Given that,

Mass of ice = 0.5 kg

Power = 2 kW = 2000 J/s

(I). If ice to completely melt to form water at 10°C

We need to calculate the heat energy

Using formula of heat

$Q=mL$

Put the value into the formula

$Q=0.5\times336000$

$Q=168000\ J$

We need to calculate the time intervals

Using formula of time

$t=\dfrac{Q}{P}$

$t=\dfrac{168000}{2000}$

$t=84\ sec$

(II). If water to attain a temperature of 100°C

We need to calculate the heat energy

Using formula of heat

$Q=mc\Delta T$

Put the value into the formula

$Q=0.5\times4200\times(100-0)$

$Q=210000\ J$

We need to calculate the time interval

Using formula of time interval

$t=\dfrac{Q}{P}$

Put the value into the formula

$t=\dfrac{210000}{2000}$

$t=105\ sec$

(III). water to change to steam at 100°C

We need to calculate the heat energy

Using formula of heat

$Q=mL$

Put the value into the formula

$Q=0.5\times2260000$

$Q=1130000\ J$

We need to calculate the time intervals

Using formula of time

$t=\dfrac{Q}{P}$

$t=\dfrac{1130000}{2000}$

$t=565\ sec$

Hence, (I). The time intervals for ice to completely melt to form water at 10°C   is 84 sec.

(II). The time intervals for water to attain a temperature of 100°C is 105 sec.

(III). The time intervals for water to change to steam at 100°C is 565 sec.