Alab 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)

The heat released for the given process is -1892 kJ

Explanation:

The processes involved in the given problem are:

Pressure is taken as constant.

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

      ......(1)

where,

q = amount of heat released = ?

m = mass of water/ice

= latent heat of fusion or vaporization

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

       .......(1)

where,

q = amount of heat released = ?

= specific heat capacity of medium

m = mass of water/ice

= final temperature

= initial temperature

Calculating the heat absorbed for each process:

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

1 kJ = 1000 J

So,

We are given:

Putting values in equation 1, we get:

We are given:

Putting values in equation 2, we get:

Total heat absorbed =

Total heat absorbed =

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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.  

the answer would be c