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27.12.2020 •

Chemistry

Please Answer ASAP!! The straighter the line on a line graph, the weaker the relationship between the two variables.

Please select the best answer from the choices provided

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Ответ:

aysenazz

05.09.2021

Chemistry

The amount of heat absorbed is $36.0\times 10^3Cal$

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:

$q=m\times \Delta H_{(f , v)}$ ......(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:

$q=m\times C_{s,l}\times (T_2-T_1)$ ......(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:

$q_1=50.0g\times 80Cal/g\\\\q_1=4000Cal$

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:

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

For process 3:

We are given:

$m=50.0g\\\Delta H_{vap}=540Cal/g$

Putting values in equation (i), we get:

$q_3=50.0g\times 540J/g\\\\q_3=27000Cal$

Calculating the total amount of heat released:

Q=q_1+q_2+q_3

$Q=[(4000)+(5000)+(27000)]Cal=36000Cal=36.0\times 10^3Cal$

Hence, the amount of heat absorbed is $36.0\times 10^3Cal$

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Please Answer ASAP!! The straighter the line on a line graph, the weaker the relationship between the two variables.

Please select the best answer from the choices provided

T
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