# Question 84a81

##### 1 Answer
Apr 29, 2017

You need to provide 156 kJ of energy.

#### Explanation:

There are five heat transfers involved in this problem.

heat to warm ice + heat to melt ice + heat to warm water + heat to boil water + heat to warm steam

$q = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$

q = mC_1ΔT_1 + mΔ_text(fus)H + mC_3ΔT_3 + mΔ_text(vap)H + mC_5ΔT_5

We can calculate each of these separately and then add them up to get the total.

q_1 = mC_1ΔT_1 = 50 color(red)(cancel(color(black)("g"))) × "2.09 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(0 -(-35))" color(red)(cancel(color(black)("°C"))) = "3658 J"

Δ_text(fus)H = "333.55 J/g"

q_2 = 50 color(red)(cancel(color(black)("g"))) × "333.55 J"/(1 color(red)(cancel(color(black)("g")))) = "16 678 J"

q_3 = mC_3ΔT_3 = 50 color(red)(cancel(color(black)("g"))) × "4.18 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(100 -0)" color(red)(cancel(color(black)("°C"))) = "20 900 J"

Δ_text(vap)H = "2260 J/g"

q_4 = 50 color(red)(cancel(color(black)("g"))) × "2260 J"/(1 color(red)(cancel(color(black)("g")))) = "113 000 J"

q_5 = mC_5ΔT_5 = 50 color(red)(cancel(color(black)("g"))) × "2.00 J"·color(red)(cancel(color(black)("g"^"-1""°C"^"-1"))) × "(128 - 100)" color(red)(cancel(color(black)("°C"))) = "2800 J"#

$q = \text{(3658 + 16 678 + 20 900 + 113 000 + 2800) J" = "156 000 J" = "156 kJ}$