Question

Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water, latent heat from ice to water 80 Cal/g)

Answer 1

Answer: The amount of heat needed is 29000 Cal.

Explanation:

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

• For process 1:

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

• For process 2:

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

• For process 3:

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.