Answer: A. 23.59 minutes.
B. 249.65 minutes
Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.
<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).
<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.
Here, <u>given that</u>:
- temperature of ice, T= -16.6°C
- rate of heat transfer,
![q=820 J.min^{-1}](https://tex.z-dn.net/?f=q%3D820%20J.min%5E%7B-1%7D%20)
- specific heat of ice,
![c_{i}= 2100 J.kg^{-1}.K^{-1}](https://tex.z-dn.net/?f=%20c_%7Bi%7D%3D%202100%20J.kg%5E%7B-1%7D.K%5E%7B-1%7D)
- latent heat of fusion of ice,
![L_{i}=334\times10^{3}J.kg^{-1}](https://tex.z-dn.net/?f=L_%7Bi%7D%3D334%5Ctimes10%5E%7B3%7DJ.kg%5E%7B-1%7D%20)
<u>Asked:</u>
1. Time require for the ice to start melting.
2. Time required to raise the temperature above freezing point.
Sol.: 1.
<u>We have the formula:</u>
![Q=mc\Delta T](https://tex.z-dn.net/?f=Q%3Dmc%5CDelta%20T)
Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.
![Q= 0.555\times2100\times16.6](https://tex.z-dn.net/?f=Q%3D%200.555%5Ctimes2100%5Ctimes16.6)
![Q= 19347.3 J](https://tex.z-dn.net/?f=Q%3D%2019347.3%20J)
Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.
∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.
![t=\frac{Q}{q}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7BQ%7D%7Bq%7D)
= 23.59 minutes.
Sol.: 2.
Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.
<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>
From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.
<u>We have the equation:</u> latent heat, ![Q_{L}= mL_{i}](https://tex.z-dn.net/?f=Q_%7BL%7D%3D%20mL_%7Bi%7D)
![Q_{L}= 0.555\times334\times10^{3}= 185370 J](https://tex.z-dn.net/?f=%20Q_%7BL%7D%3D%200.555%5Ctimes334%5Ctimes10%5E%7B3%7D%3D%20185370%20J)
<u>Now the time required for supply of 185370 J:</u>
![t=\frac{Q_{L}}{q}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7BQ_%7BL%7D%7D%7Bq%7D)
![t=\frac{185370}{820}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B185370%7D%7B820%7D)
t= 226.06 minutes
∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59
= 249.65 minutes