Respuesta :
Answer : It takes less amount of heat to metal 1.0 Kg of ice.
Solution :
The process involved in this problem are :
[tex](1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)[/tex]
Now we have to calculate the amount of heat released or absorbed in both processes.
For process 1 :
[tex]Q_1=m\times \Delta H_{fusion}[/tex]
where,
[tex]Q_1[/tex] = amount of heat absorbed = ?
m = mass of water or ice = 1.0 Kg
[tex]\Delta H_{fusion}[/tex] = enthalpy change for fusion = [tex]3.35\times 10^5J/Kg[/tex]
Now put all the given values in [tex]Q_1[/tex], we get:
[tex]Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J[/tex]
For process 2 :
[tex]Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})[/tex]
where,
[tex]Q_2[/tex] = amount of heat absorbed = ?
m = mass of water = 1.0 Kg
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4186J/Kg^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]0^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]100^oC[/tex]
Now put all the given values in [tex]Q_2[/tex], we get:
[tex]Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC[/tex]
[tex]Q_2=4.186\times 10^5J[/tex]
From this we conclude that, [tex]Q_1<Q_2[/tex] 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.
