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.

Step-by-step 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)* (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* 0.5Cal/g^oC* (0-(-30))^oC=3000Cal`

• For process 2:

`q_2=m* 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* 80Cal/g=16000Cal`

• For process 3:

`q_3=m* C_(p,l)* (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* 1Cal/g^oC* (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.