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

Step-by-step 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* C_(p,m)* (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* 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* 2.03J/g^oC* (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,
`((6.01kJ)/(1mol))* ((1000J)/(1kg))* ((1mol)/(18g))=334J/g`

We are given:

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

Putting values in equation 2, we get:

`q_2=856g* 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* 4.184J/g^oC* (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,
`((40.79kJ)/(1mol))* ((1000J)/(1kg))* ((1mol)/(18g))=2266J/g`

We are given:

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

Putting values in equation 2, we get:

`q_4=856g* 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* 1.99J/g^oC* (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