Answer to Question #302163 in General Chemistry for Cherry

Question

How much heat is required to convert 40.0 grams of ice at -10.0 degrees Celsius to 100.0 degrees Celsius? The specific heat of ice is 2.1 J/g degrees Celsius.

Answer 1

We will need the following useful information to solve this question:

heat of fusion of water = 334 J/g

heat of vaporization of water = 2257 J/g

specific heat of ice = 2.09 J/g·°C

specific heat of water = 4.18 J/g·°C

specific heat of steam = 2.09 J/g·°C

We need to calculate the sum total of all the heat energies required to;

I. To heat ice from -100^oC to 0^oC

II. To melt ice from 0^oC ice to 0^oC water

II. To heat water from 0^oC to 100^oC

I. To heat ice from -100^oC to 0^oC

$q=mC\Delta T$

m = 40.0g

C= 2.1J/g^oC

$\Delta$T = 0^oC - (-100^oC) =100^oC

q = 40.0g x 2.1J/g^oC x 100^oC

= 8400J

II. To melt ice from 0^oC ice to 0^oC water

q = m$\Delta$H_f

where $\Delta$Hf = 334j/g

q = 4.0g x 334J/g

= 13360J

II. To heat water from 0^oC to 100^oC

q = mC$\Delta$T

$\Delta$T = (100 - 0)^oC

= 4.0g x 4.18J/g^oC x 100^oC

= 16720J

Heat_Total = Heat_{Step 1} + Heat_{Step 2} + Heat_{Step 3}

= (8400 + 13360 + 16720) J

= 38480J

= 38.48kJ