Question

How much heat is required to change 56.0 g of ice (H2O) at 263 K to vapor (steam) at 400 K?

Answer 1

Heat required = 173 kJ

Further explanation

Given

56.0 g of ice

Temperatur at 263 K(-10 C) to 400 K(127 C)

Required

Heat needed

Solution

1. raise the temperature(-10 C to 0 C)⇒c ice=2.09 J/g C

$\tt Q=56\times 2.09\times (0-(-10)=1170.4~J$

2. phase change (ice to water)⇒Heat of fusion water=334 J/g

$\tt Q=56\times 334=18704~J$

3. raise the temperature(0 C to 100 C)⇒c water= 4.18 J/g C

$\tt Q=56\times 4.18\times (100-0)=23408~J$

4. phase change(water to vapor)⇒heat of vaporization water=2260 J/g

$\tt Q=56\times 2260=126560~J$

5. raise the temperature(100 C to 127 C)⇒c vapor=2.09 J/g C

$\tt Q=56\times 2.09\times (127-100)=3160.08~J$

Total heat :

1170.4+18704+23408+126560+3160.08=173,002.48 J=173 kJ