Answer: The amount of heat needed is 29000 Cal.
Explanation:
The process involved in this problem are:
[tex](1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)[/tex]
Now, we calculate the amount of heat released or absorbed in all the processes.
[tex]q_1=mC_{p,s}\times (T_2-T_1)[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
m = mass of ice = 200 g
[tex]T_2[/tex] = final temperature = [tex]0^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]-30^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal[/tex]
[tex]q_2=m\times L_f[/tex]
where,
[tex]q_1[/tex] = amount of heat absorbed = ?
m = mass of water or ice = 200 g
[tex]L_f[/tex] = latent heat of fusion = 80 Cal/g
Putting all the values in above equation, we get:
[tex]q_2=200g\times 80Cal/g=16000Cal[/tex]
[tex]q_3=m\times C_{p,l}\times (T_{2}-T_{1})[/tex]
where,
[tex]q_3[/tex] = amount of heat absorbed = ?
m = mass of water = 200 g
[tex]T_2[/tex] = final temperature = [tex]50^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]0^oC[/tex]
Putting all the values in above equation, we get:
[tex]q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal[/tex]
Calculating the total heat absorbed, we get:
[tex]Q=q_1+q_2+q_3[/tex]
[tex]Q=3000+16000+10000=29000Cal[/tex]
Hence, the amount of heat needed is 29000 Cal.
