Question

Calculate the total amount of energy required in calories to convert 50.0 g of ice at 0.00 degrees Celsius to steam at 100. degrees Celsius.

Specific heat capacity of water is 1.00 cal/g OC

Hfusion = 80 cal/g OC and Hvap = 540 cal/g OC

Write the complete equation you will use.

Substitute the values in the equation in step 1.

Report the math answer with 3 sig figs and the correct unit.

Answer 1

Answer: The amount of heat absorbed is
`36.0* 10^3Cal`

Step-by-step explanation:

Few processes involved are:

(1):
`H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)`

(2):
`H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)`

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

Calculating the heat absorbed for the process having same temperature:

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

where,

q is the amount of heat absorbed, m is the mass of sample and 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=50.0g\\\Delta H_(fusion)=80Cal/g`

Putting values in equation (i), we get:

`q_1=50.0g* 80Cal/g\\\\q_1=4000Cal`

• For process 2:

We are given:

`m=50.0g\\C=1.00Cal/g^oC\\T_2=100^oC\\T_1=0^oC`

Putting values in equation (i), we get:

`q_2=50g* 1Cal/g^oC* (100-0)\\\\q_2=5000Cal`

• For process 3:

We are given:

`m=50.0g\\\Delta H_(vap)=540Cal/g`

Putting values in equation (i), we get:

`q_3=50.0g* 540J/g\\\\q_3=27000Cal`

Calculating the total amount of heat released:

`Q=q_1+q_2+q_3`

`Q=[(4000)+(5000)+(27000)]Cal=36000Cal=36.0* 10^3Cal`

Hence, the amount of heat absorbed is
`36.0* 10^3Cal`