1. THERMODYNAMICS: MATH PRESENTATION

2. EXAMPLE 1: What quantity of heat is required to raise the temperature of 450 grams of water from 15°C to 85°C? The specific heat capacity of water is 4.18 J/g/°C. G: m = 450 g C = 4.18 J/g/°C T initial = 15°C T final = 85°C U: We wish to determine the value of Q - the quantity of heat. E: To do so, we would use the equation Q = mCΔT. S: T = T final - T initial = 85°C - 15°C = 70.°C Q = mCΔT = (450 g)(4.18 J/g/°C)(70.°C) S: Q = 131670 J Q = 1.3x10 5 J = 130 kJ

3. EXAMPLE 2: A 12.9 gram sample of an unknown metal at 26.5°C is placed in a Styrofoam cup containing 50.0 grams of water at 88.6°C. The water cools down and the metal warms up until thermal equilibrium is achieved at 87.1°C. Assuming all the heat lost by the water is gained by the metal and that the cup is perfectly insulated, determine the specific heat capacity of the unknown metal. The specific heat capacity of water is 4.18 J/g/°C. Part 1: Determine the Heat Lost by the Water

4. EXAMPLE 2: Part 1: Determine the Heat Lost by the Water Given: m = 50.0 g C = 4.18 J/g/°C T initial = 88.6°C T final = 87.1°C ΔT = -1.5°C (T final - T initial ) Unknown: Q water : Q water = mCΔT = (50.0 g)(4.18 J/g/°C)(-1.5°C) Q water = -313.5 J (unrounded) (The - sign indicates that heat is lost by the water)

5. EXAMPLE 2: Part 2: Determine the value of C metal Given: Q metal = 313.5 J (use a + sign since the metal is gaining heat) m = 12.9 g T initial = 26.5°C T final = 87.1°C ΔT = (T final - T initial ) Unknown: Solve for C metal : Steps: Rearrange Q metal = m metal C metalΔT metal to obtain C metal = Q metal / (m metalΔT metal ) C metal = Q metal / (m metalΔT metal ) = (313.5 J)/[(12.9 g)(60.6°C)] Solution: C metal = 0.40103 J/g/°C C metal = 0.40 J/g/°C (rounded to two significant digits)

6. EXAMPLE 3: Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g. Givens: mass = 48.2 grams heat of fusion = 333 J/g Equation: Q = mΔH fusion Steps: Q = mΔH fusion = (48.2 g)(333 J/g) Solution: Q = 16050.6 J Q = 1.61 x 10 4 J = 16.1 kJ

7. EXAMPLE 4: What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g.

8. EXAMPLE 4: Givens: Given Info about Ice: m = 50.0 g ΔH fusion = 333 J/g Given Info about Liquid Water: C = 4.18 J/g/°C T initial = 26.5°C T final = 0.0°C ΔT = -26.5°C (T final - T initial ) Equation: The energy gained by the ice is equal to the energy lost from the water. Q ice = -Q liquid water The - sign indicates that the one object gains energy and the other object loses energy. We can calculate the left side of the above equation as follows:

9. EXAMPLE 4: Steps: Q ice = mΔH fusion = (50.0 g)(333 J/g) Q ice = 16650 J Now we can set the right side of the equation equal to mCΔT and begin to substitute in known values of C and ΔT in order to solve for the mass of the liquid water. The solution is: 16650 J = -Q liquid water 16650 J = -m liquid water C liquid water ΔT liquid water 16650 J = -m liquid water (4.18 J/g/°C)(-26.5°C) 16650 J = -m liquid water (-110.77 J/°C) Solution: m liquid water = -(16650 J)/(-110.77 J/°C) m liquid water = 150.311 g m liquid water = 1.50x10 2 g (rounded to three significant digits)