Question

Calculate the total energy, in kilojoules, that is needed to turn a 46 g block

of ice at -25 degrees C into water vapor at 100 degrees C.

Answer 1

Answer: The amount of heat absorbed is 141.004 kJ.

Step-by-step explanation:

In order to calculate the amount of heat released while converting given amount of steam (gaseous state) to ice (solid state), few processes are involved:

(1):
`H_2O (s) (-25^oC, 248K) \rightleftharpoons H_2O(s) (0^oC,273K)`

(2):
`H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)`

(3):
`H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)`

(4):
`H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)`

Calculating the heat absorbed for the process having the same temperature:

`q=m* \Delta H_((f , v))` ......(i)

where,

q is the amount of heat absorbed, m is the mass of sample and
`\Delta H_((f , v))` is the enthalpy of fusion or vaporization

Calculating the heat released for the process having different temperature:

`q=m* C_(s,l)* (T_2-T_1)` ......(ii)

where,

`C_(s,l)` = specific heat of solid or liquid

`T_2\text{ and }T_1` are final and initial temperatures respectively

• For process 1:

We are given:

`m=46g\\C=2.108J/g^oC\\T_2=0^oC\\T_1=-25^oC`

Putting values in equation (i), we get:

`q_1=46g* 2.108J/g^oC* (0-(-25))\\\\q_1=2424.2J`

• For process 2:

We are given:

`m=46g\\\Delta H_(fusion)=334J/g`

Putting values in equation (i), we get:

`q_2=46g* 334J/g\\\\q_2=15364J`

• For process 3:

We are given:

`m=46g\\C=4.186J/g^oC\\T_2=100^oC\\T_1=0^oC`

Putting values in equation (i), we get:

`q_3=46g* 4.186J/g^oC* (100-0)\\\\q_3=19255.6J`

• For process 4:

We are given:

`m=46g\\\Delta H_(vap)=2260J/g`

Putting values in equation (i), we get:

`q_4=46g* 2260J/g\\\\q_4=103960J`

Calculating the total amount of heat released:

`Q=q_1+q_2+q_3+q_4`

`Q=[(2424.2)+(15364)+(19255.6)+(103960)]`

`Q=141003.8J=141.004kJ` (Conversion factor: 1 kJ = 1000J)

Hence, the amount of heat absorbed is 141.004 kJ.