Sample Problem solutions Thermal Energy Sample Problem solutions

1. 1 kg of water (specific heat = 4184 J/(kg K)) is heated from freezing (0°C) to boiling (100°C). What is the change in thermal energy?

Q = mCP Dt Q = (1kg)(4184 J/(kg K)) (100 – 0) Q = 418400 J or 418.4 kJ

2. A 100 g block of metal absorbs 900 Joules of thermal energy when its temperature changes from 20oC to 40oC. Calculate the specific heat of the metal and use your specific heat table to determine its identity.

Q = mCP Dt 900 J = (0.1kg) CP (40 – 20) CP = 450 J/(kg K)  Iron/Steel

3. In the morning, a concrete statue has a temperature of 15°C. During the day, it sits in the sun and warms up to 40°C, gaining 90,000 J of thermal energy. What is its mass?

Q = mCP Dt 90000 J = m (850 J/(kg K)) (40 – 15) m = 4.24 kg

4. How much heat is required to melt 5kg if ice?

Q = mL Q = (5 kg) (334000 J/kg) Q = 1670000 J or 1670 kJ

5. How much thermal energy is released when 10kg of copper cools from 195°C to 15°C?

Q = mCP Dt Q = (10 kg) (390 J/(kg K)) (15 – 195) Q = -702000 J or -702 kJ

6. How much thermal energy is required to completely melt 4.0kg of Aluminum?

Q = mL Q = (4 kg) (396000) Q = 1584000 J or 1584kJ

If you were to mix 2. 0kg of ice (0°C) and 4 If you were to mix 2.0kg of ice (0°C) and 4.0kg of water at 35°C what will be the final temperature and state of the mixture once it reaches thermal equilibrium?

To melt 2kg of ice, we need this much energy: Q = 2kg x 334000 J Q = 668000 J Can we get this much heat energy from 4kg of water at 35o C? Best we can get is Q = mCpDt Q = 4kg (4186) (0 – 35) Q = -586040 J Since 668000 J is greater than 586040 J, the ice did not all melted meaning there is ice left in our mix. If ice is present, then the temperature is 0 degrees.

To be more specific about our solution, we can calculate how much ice remain in our mix. Taking all the thermal energy from the water to 0o C, we got 586040 J. All of this energy went into melting the ice, so how much ice did we melt. Q = mL 586040 = m (334000 J) m = 1.75 kg Meaning we have 2 kg – 1.75 kg left of ice. 0.25 kg of ice remain and 5.75 kg of water.

For cooling of metals and other materials that does not need to go through phase change, we can use the following equation: Qgain = Qlost mCP (Tfinal – Tinitial) = mCP (Tinitial – Tfinal)

Thermal Energy Additional Practice 1. A 500 g glass dish is heated from 25oC to 350oC. How much thermal energy in kilojoules does it absorb during the heating?

Q = mCpDt Q = (0.5 kg)(670 J/(kg K))(350 – 25) = 108875 J = 108.875 kJ

2. A sample of water absorbs 50 kJ of energy when it is heated from 20oC to 60oC. What is the mass of the water?

Q = mCpDt 50000 = m(4186 J/(kg K))(60 – 20) m = 0.299 kg

A 100 g block of metal absorbs 900 Joules of thermal energy when its temperature changes from 20oC to 40oC. Calculate the specific heat of the metal, and then use your specific heat chart to determine the identity of the metal.

Q = mCP Dt 900 J = (0.1kg) CP (40 – 20) CP = 450 J/(kg K)  Iron/Steel

4. How much thermal energy is removed from 300 grams of steam at 115oC to condense it to water at 15oC?

Cooling the steam from 115o C to 100o C. Q = mCP Dt = (0 Cooling the steam from 115o C to 100o C. Q = mCP Dt = (0.3 kg)(2000 J/(kg K))(100o C – 115o C) = -9000 J Condensing the steam at 100o C Q = mL = (0.3 kg)(2256000 J/kg) = -676800 J Cooling water from 100o C to 15o C = (0.3 kg)(4186 J/(kg K))(15o C – 100o C) = -106743 J Total Thermal Energy = -9000 + -676800 + -106743 = -792543 J

5. How much thermal energy is needed to change 25 grams of steam at 130oC to ice at -15oC.

Cooling the steam from 130o C to 100o C. Q = mCP Dt = (0 Cooling the steam from 130o C to 100o C. Q = mCP Dt = (0.025 kg)(2000 J/(kg K))(100o C – 130o C) = -1500 J Condensing the steam at 100o C Q = mL = (0.025 kg)(2256000 J/kg) = -56400 J Cooling water from 100o C to 0o C = (0.025 kg)(4186 J/(kg K))(0o C – 100o C) = -10465 J Freezing the water = (0.025 kg)(334000 J/kg) = -8350 J Cooling ice = (0.025 kg)(2100 J/(kg K))(-15o C – 0o C) = -787.5 J Total = -77502.5 J

6. A 500 g sample of water at 50oC is mixed with another 500 g sample of water at 70oC. What is the final temperature of the mixture? Hint: this problem is very short!

Since the two samples have the same mass and are the same materials, it is just the average of the two temperature (50 + 70)/ 2 = 60o C

7. A 100 g sample of gold at 200oC is placed in 300 g of water at 25oC 7. A 100 g sample of gold at 200oC is placed in 300 g of water at 25oC. What is the final temperature of the gold and water?

Qgain = Qlost mCP (Tfinal – Tinitial) = mCP (Tinitial – Tfinal) (0 Qgain = Qlost mCP (Tfinal – Tinitial) = mCP (Tinitial – Tfinal) (0.3 kg)(4186 J/kg K)(Tfinal – 25) = (0.1 kg)(129 J/kg K)(200 – Tfinal) 1255.8 Tfinal – 31395 = 2580 – 12.9 Tfinal 1268.7 Tfinal = 33975 Tfinal = 26.78

8. A 10 g sample of an unknown substance at 100oC is placed into a 10 g sample of water at 10oC. The final temperature of the substance and water is 14.8oC. What is the specific heat of the unknown substance? Use your specific heat chart to determine the identity of the substance.

Qgain = Qlost mCP (Tfinal – Tinitial) = mCP (Tinitial – Tfinal)