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# Calculate the amount of heat needed to change 45.0 grams of ice at -25.0 °C to steam at 250.0 °C. ?

Full question: Calculate the amount of heat needed to change 45.0 grams of ice at -25.0 °C to steam at 250.0 °C. (For water, ΔHfus = 333 J/g; ΔHvap = 2260 J/g; Specific Heat (ice) = 2.10 J g-1 K-1; Specific Heat (steam) = 2.00 J g-1 K-1)

### 1 Answer

- 4 months agoFavorite Answer
First, heat the ice up to 0 C. A change of 1 degree Celsius is the same as the change of 1 degree Kelvin

2.10 J/(g * K) * 45 g * 25 K =>

25 * 45 * 2.10 Joules

25 * 45 * 2.1 Joules =>

(100/4) * 45 * 2.1 Joules =>

45 * 210 / 4 Joules =>

90 * 210 / 8 Joules =>

18900 / 8 Joules >

9450 / 4 Joules =>

4725 / 2 Joules =>

2362.5 Joules

Now calculate how much energy it would take to melt the ice

45 g * 333 J/g =>

5 * 9 * 333 Joules =>

5 * (2700 + 270 + 27) Joules =>

5 * 2997 Joules =>

5 * (3000 - 3) Joules =>

15000 - 15 Joules =>

14985 Joules

Now calculate how much energy it would take to heat the water up to 100 C

4.186 J/(g * K) * 100 K * 45 g =>

45 * 4.186 * 100 Joules =>

45 * 418.6 Joules =>

90 * 209.3 Joules =>

9 * 2093 Joules =>

18000 + 810 + 27 Joules =>

18837 Joules

Now calculate how much energy it would take to turn the water into steam

45 g * 2260 J/g =>

45 * 2260 Joules =>

90 * 1130 Joules =>

9 * 113 * 100 Joules =>

(900 + 90 + 27) * 100 Joules =>

(900 + 117) * 100 Joules =>

1017 * 100 Joules =>

101700 Joules

Now calculate how much energy it would take to heat the steam another 150 degrees

45 g * 2 J/(g * K) * 150 K =>

45 * 2 * 150 Joules =>

90 * 150 Joules =>

13500 Joules

Add them all together

2362.5 Joules + 14985 Joules + 18837 Joules + 101700 Joules + 13500 Joules =>

151,384.5 Joules

To 3 sf

151000 Joules

To 4 sf

151400 Joules