Question

At 1 atm, how much energy is required to heat 53.0 g of H2O(s) at –18.0 °C to H2O(g) at 143.0 °C?

Answer 1

Given:

m H2O = 53.0 g , Initial T = -18.0⁰ C , final T = 143.0 ⁰C

Solution:

There are following transitions take place while going from – 18.0 ⁰ C to 143.0

First : -18.0 ⁰C to 0 ⁰C

q1 = m C delta T

m is mass in g, C is specific heat capacity of ice water. , Delta T = change in T ( T final – T initial )

q is change in heat.

Heat capacity C of ice = 2.03 J / g ⁰ C

Lets use given values to calculated q1

q 1 = 53.0 g x 2.03 J / g ⁰ C x ( 0 – ( -18.0 ))⁰C

= 1936.62 J

Second :

From 0 to 0 ⁰ C

Here solid water converts to liquid water and so we use delta H of melting/heat of fusion.

q 2 = m x delta Hf

m is mass in g, Delta Hf is heat of fusion.

Delta Hf of water = 333.55 J / g

q 2 = 53.0 g x 333.55 J / g

= 17678.15 J

Third :

From : 0 to 100 ⁰ C

q 3 = m C delta T

m is mass in g, C is specific heat capacity of liquid water , Delta T = change in T ( T final – T initial )

q is change in heat.

q 3 = 53.0 g x 4.184 J / g ⁰ C x (100 – 0 ) ⁰C

=22175.2 J

Fourth : Fro 100 to 100 : heat of vaporization

q 4 = m x Delta Hv

m is mass in g , Delta Hf is heat of vaporization.

q 4 = 53.0 g x 2260 J/g

= 119780 J

Fifth : from 100 ⁰ to 143.0 ⁰C

Here we use heat capacity of vapor (water)

q 5 = 53.0 g x 2.02 J / g ⁰C x ( 143.0 – 100 ) ⁰C

= 4603.58 J

Lets add these values together.

q ( total heat required) = sum of all q’s

= (1936.62 + 17678.15+22175.2 + 119780+4603.58 ) J

= 166173.6 J

Energy required for this conversion = 166173.6 J

= 166.2 kJ