Answer : It takes less amount of heat to metal 1.0 Kg of ice.
Solution :
The process involved in this problem are :
![(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)](https://tex.z-dn.net/?f=%281%29%3AH_2O%28s%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%280%5EoC%29%5C%5C%5C%5C%282%29%3AH_2O%28l%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%28100%5EoC%29)
Now we have to calculate the amount of heat released or absorbed in both processes.
<u>For process 1 :</u>
![Q_1=m\times \Delta H_{fusion}](https://tex.z-dn.net/?f=Q_1%3Dm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D)
where,
= amount of heat absorbed = ?
m = mass of water or ice = 1.0 Kg
= enthalpy change for fusion = ![3.35\times 10^5J/Kg](https://tex.z-dn.net/?f=3.35%5Ctimes%2010%5E5J%2FKg)
Now put all the given values in
, we get:
![Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J](https://tex.z-dn.net/?f=Q_1%3D1.0Kg%5Ctimes%203.35%5Ctimes%2010%5E5J%2FKg%3D3.35%5Ctimes%2010%5E5J)
<u>For process 2 :</u>
![Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})](https://tex.z-dn.net/?f=Q_2%3Dm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29)
where,
= amount of heat absorbed = ?
m = mass of water = 1.0 Kg
= specific heat of liquid water = ![4186J/Kg^oC](https://tex.z-dn.net/?f=4186J%2FKg%5EoC)
= initial temperature = ![0^oC](https://tex.z-dn.net/?f=0%5EoC)
= final temperature = ![100^oC](https://tex.z-dn.net/?f=100%5EoC)
Now put all the given values in
, we get:
![Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC](https://tex.z-dn.net/?f=Q_2%3D1.0Kg%5Ctimes%204186J%2FKg%5EoC%5Ctimes%20%28100-0%29%5EoC)
![Q_2=4.186\times 10^5J](https://tex.z-dn.net/?f=Q_2%3D4.186%5Ctimes%2010%5E5J)
From this we conclude that,
that means it takes less amount of heat to metal 1.0 Kg of ice.
Hence, the it takes less amount of heat to metal 1.0 Kg of ice.