Question

A lab scientist cools a liquid sample of water (2.6 kg) at 0.00°C to -192°C. The water turns to ice as this temperature change occurs. How much heat is released during this process? [For water, Lf = 334 kJ/kg and LV = 2257 kJ/kg. The specific heat for ice is 2050 J/(kg·K)]. Report your answer in kJ. (round your answer to a whole number - no decimal places)

Answer 1

Answer: The heat released for the given process is -1892 kJ

Step-by-step explanation:

The processes involved in the given problem are:

`1.)H_2O(l)(0^oC,273K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(s)(-192^oC,81K)`

Pressure is taken as constant.

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

`q=m* L_(f,v)` ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

`L_(f,v)` = latent heat of fusion or vaporization

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

`q=m* C_(p,m)* (T_(2)-T_(1))` .......(1)

where,

q = amount of heat released = ?

`C_(p,m)` = specific heat capacity of medium

m = mass of water/ice

`T_2` = final temperature

`T_1` = initial temperature

Calculating the heat absorbed for each process:

• For process 1:

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

1 kJ = 1000 J

So,
`((-334kJ)/(1kg))* ((1000J)/(1kJ))=-334* 10^3J/kg`

We are given:

`m=2.6kg\\L_f=-334* 10^3J/kg`

Putting values in equation 1, we get:

`q_1=2.6kg* (-334* 10^3J/kg)=-868400J`

• For process 2:

We are given:

`m=2.6kg\\C_(p,s)=2050J/kg.K\\T_1=273K\T_2=81K`

Putting values in equation 2, we get:

`q_2=2.6kg* 2050J/kg.K* (81-(273))^oC\\\\q_2=1023360J`

Total heat absorbed =
`q_1+q_2`

Total heat absorbed =
`[-868400+(-1023360)]J=-1891760J=-1891.76kJ\approx -1892kJ`

Hence, the heat released for the given process is -1892 kJ