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

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:

$q=m\times L_{f,v}$       ......(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:

$q=m\times C_{p,m}\times (T_{2}-T_{1})$        .......(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, $(\frac{-334kJ}{1kg})\times (\frac{1000J}{1kJ})=-334\times 10^3J/kg$

We are given:

$m=2.6kg\\L_f=-334\times 10^3J/kg$

Putting values in equation 1, we get:

$q_1=2.6kg\times (-334\times 10^3J/kg)=-868400J$

• 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:

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

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