Question

How much heat (in kJ) is needed to convert 856 g of ice at −10.0°C to steam at 126.0°C? (The specific heats of ice, water, and steam are 2.03 J/g · °C, 4.184 J/g · °C, and 1.99 J/g · °C, respectively. The heat of fusion of water is 6.01 kJ/mol, the heat of vaporization is 40.79 kJ/mol.)

Answer 1

Answer: The heat required for the given process is 2659.3 kJ

Explanation:

The processes involved in the given problem are:  

$1.)H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\2.)H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\3.)H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)\\4.)H_2O(l)(100^oC)\rightarrow H_2O(g)(100^oC)\\5.)H_2O(g)(100^oC)\rightarrow H_2O(g)(126^oC)$

Pressure is taken as constant.

To calculate the amount of heat absorbed at different temperature, we use the equation:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$        .......(1)

where,

q = amount of heat absorbed = ?

$C_{p,m}$ = specific heat capacity of medium

m = mass of water/ice

$T_2$ = final temperature

$T_1$ = initial temperature

To calculate the amount of heat released at same temperature, we use the equation:

$q=m\times L_{f,v}$       ......(2)

where,

q = amount of heat absorbed = ?

m = mass of water/ice

$L_{f,v}$ = latent heat of fusion or vaporization

Calculating the heat absorbed for each process:

• For process 1:

We are given:

$m=856g\\C_{p,s}=2.03J/g^oC\\T_1=-18^oC\\T_2=0^oC$

Putting values in equation 1, we get:

$q_1=856\times 2.03J/g^oC\times (0-(-18))^oC\\\\q_1=31278.24J$

• For process 2:

Converting the latent heat of fusion in J/g, we use the conversion factor:

Molar mass of water = 18 g/mol

1 kJ = 1000 J

So, $(\frac{6.01kJ}{1mol})\times (\frac{1000J}{1kg})\times (\frac{1mol}{18g})=334J/g$

We are given:

$m=856g\\L_f=334J/g$

Putting values in equation 2, we get:

$q_2=856g\times 334J/g=285904J$

• For process 3:

We are given:

$m=856g\\C_{p,l}=4.184J/g^oC\\T_1=0^oC\\T_2=100^oC$

Putting values in equation 1, we get:

$q_3=856g\times 4.184J/g^oC\times (100-(0))^oC\\\\q_3=358150.4J$

• For process 4:

Converting the latent heat of vaporization in J/g, we use the conversion factor:

Molar mass of water = 18 g/mol

1 kJ = 1000 J

So, $(\frac{40.79kJ}{1mol})\times (\frac{1000J}{1kg})\times (\frac{1mol}{18g})=2266J/g$

We are given:

$m=856g\\L_v=2266J/g$

Putting values in equation 2, we get:

$q_4=856g\times 2266J/g=1939696J$

• For process 5:

We are given:

$m=856g\\C_{p,g}=1.99J/g^oC\\T_1=100^oC\\T_2=126^oC$

Putting values in equation 1, we get:

$q_5=856g\times 1.99J/g^oC\times (126-(100))^oC\\\\q_5=44289.44J$

Total heat absorbed = $q_1+q_2+q_3+q_4+q_5$

Total heat absorbed = $[31278.24+285904+358150.4+1939696+44289.44]J=2659318.08J=2659.3kJ$

Hence, the heat required for the given process is 2659.3 kJ