Respuesta :
Answer: The amount of heat absorbed is [tex]36.0\times 10^3Cal[/tex]
Explanation:
Few processes involved are:
(1): [tex]H_2O (s) (0^oC, 273K) \rightleftharpoons H_2O(l) (0^oC,273K)[/tex]
(2): [tex]H_2O (l) (0^oC, 273K) \rightleftharpoons H_2O(l) (100^oC,373K)[/tex]
(3): [tex]H_2O (l) (100^oC, 373K) \rightleftharpoons H_2O(g) (100^oC,373K)[/tex]
Calculating the heat absorbed for the process having same temperature:
[tex]q=m\times \Delta H_{(f , v)}[/tex] ......(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:
[tex]q=m\times C_{s,l}\times (T_2-T_1)[/tex] ......(ii)
where,
[tex]C_{s,l}[/tex] = specific heat of solid or liquid
[tex]T_2\text{ and }T_1[/tex] are final and initial temperatures respectively
- For process 1:
We are given:
[tex]m=50.0g\\\Delta H_{fusion}=80Cal/g[/tex]
Putting values in equation (i), we get:
[tex]q_1=50.0g\times 80Cal/g\\\\q_1=4000Cal[/tex]
- For process 2:
We are given:
[tex]m=50.0g\\C=1.00Cal/g^oC\\T_2=100^oC\\T_1=0^oC[/tex]
Putting values in equation (i), we get:
[tex]q_2=50g\times 1Cal/g^oC\times (100-0)\\\\q_2=5000Cal[/tex]
- For process 3:
We are given:
[tex]m=50.0g\\\Delta H_{vap}=540Cal/g[/tex]
Putting values in equation (i), we get:
[tex]q_3=50.0g\times 540J/g\\\\q_3=27000Cal[/tex]
Calculating the total amount of heat released:
[tex]Q=q_1+q_2+q_3[/tex]
[tex]Q=[(4000)+(5000)+(27000)]Cal=36000Cal=36.0\times 10^3Cal[/tex]
Hence, the amount of heat absorbed is [tex]36.0\times 10^3Cal[/tex]
