How much thermal energy is added to 10.0 g of ice at −20.0°c to convert it to water vapor at 120.0°c?

7479 cal.

31262.2 joules

Explanation:

This is a calorimetry problem where water in its three states changes from ice to vapor.

We must use, the calorimetry formula and the formula for latent heat.

Q = m . C . ΔT

Q = Clat . m

First of all, let's determine the heat for ice, before it melts.

10 g . 0.5 cal/g°C ( 0° - (-20°C) = 100 cal

Now, the ice has melted.

Q = Clat heat of fusion . 10 g

Q = 79.7 cal/g . 10 g → 797 cal

We have water  at 0°, so this water has to receive heat until it becomes vapor. Let's determine that heat.

Q = m . C . ΔT

Q = 10 g . 1 cal/g°C (100°C - 0°C) → 1000 cal

Water is ready now, to become vapor so let's determine the heat.

Q = Clat heat of vaporization . m

Q = 539.4 cal/g . 10 g → 5394 cal

Finally we have vapor water, so let's determine the heat gained when this vapor changes the T° from 100°C to 120°

Q = m . C . ΔT

Q = 10 g . 0.470 cal/g°C . (120°C - 100°C) → 94 cal

Now, we have to sum all the heat that was added in all the process.

100 cal + 797 cal + 1000 cal + 5394 cal + 94 cal =7479 cal.

We can convert this unit to joules, which is more acceptable for energy terms.

1 cal is 4.18 Joules.

Then, 7479 cal are (7479 . 4.18) = 31262.2 joules

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).

water at room temperature is flowing through a pipe with pressure taps located 10 ft apart along one section of the pipe, which are connected to a mercury (s=13.6) manometer.