Question

$40g$ of ice at $-10°\mathrm{C}$ is heated by a heater of power $250\mathrm{W}$, such that water formed from it, attains the temperature equal to the boiling point of water. For how long is the heater switched on? [S.H.C. of ice$=2{\mathrm{Jg}}^{-1}{\mathrm{C}}^{-1}$; Sp. latent heat of ice$=340{\mathrm{Jg}}^{-1}$]

Answer 1

Step 1: Given data

Mass of ice is $m=40g$at $-10°\mathrm{C}$

Power of heater$=250W=250J/s$.

Specific Heat Capacity of ice is $c_1=2{\mathrm{Jg}}^{-1}{\mathrm{C}}^{-1}$.

Specific latent heat of ice is $L=340{\mathrm{Jg}}^{-1}$.

Step 2: Finding the heat required to convert ice at $-10°C$ to $0°C$

Let the amount of heat required to convert ice at $-10°C$ to $0°C$ be $H_1$

So, $H_1=m{c}_1{\theta }_R$.

Substituting the known values, we get

$H_1=40\times 2\times [0-(-10\left)\right]=40\times 2\times 10=800\mathrm{J}$

Step 3: Finding the heat required to convert ice at $0°C$ to water at $0°C$

Let the amount of heat required to convert ice at $0°C$ to water $0°C$ be $H_2$

So, $H_2=mL$.

Putting the values, we get

$H_2=40\times 340=13600\mathrm{J}$

Step 4: Finding the heat required to convert water at $0°C$to water at $100°C$

Let the amount of heat required to convert water at $0°C$to water at $100°C$ be $H_3$

Specific heat capacity of water is $c_2=4.2J/g°C$.

So, $H_3=m{c}_2{\theta }_R$.

Putting the values, we get

$H_3=40\times 4.2\times (100-0)=40\times 4.2\times 100=16800\mathrm{J}$

Step 5: Finding the total amount of heat required

Total amount of heat is $H=H_1+H_2+H_3=800+13600+16800=31200\mathrm{J}$.

Step 6: Finding the time for which the heater was switched ON

Now, the time for the heater is$\text{Time(t) = Heat consumed(H)}/\text{Power (P)}$.

Putting the known values,

$t=31200/250=124.8\mathrm{s}$

Hence, the heater is switched on for $124.8$ seconds.