THERMODYNAMICS:MATH PRESENTATION

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• Tinitial = 15°C

• Tfinal = 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 = m•C•ΔT.

• S:• T = Tfinal - Tinitial = 85°C - 15°C = 70.°C

• Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C)

• S:• Q = 131670 J• Q = 1.3x105 J = • 130 kJ

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

EXAMPLE 2:

• Part 1: Determine the Heat Lost by the Water• Given: • m = 50.0 g• C = 4.18 J/g/°C • Tinitial = 88.6°C

• Tfinal = 87.1°C

• ΔT = -1.5°C (Tfinal - Tinitial)

• Unknown: Qwater:

• Qwater = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)

• Qwater = -313.5 J (unrounded)

• (The - sign indicates that heat is lost by the water)

EXAMPLE 2:

• Part 2: Determine the value of Cmetal

• Given: • Qmetal = 313.5 J (use a + sign since the metal is gaining heat)

• m = 12.9 g• Tinitial = 26.5°C

• Tfinal = 87.1°C

• ΔT = (Tfinal - Tinitial )

• Unknown: Solve for Cmetal:

• Steps: • Rearrange Qmetal = mmetal•Cmetal•ΔTmetal to obtain Cmetal = Qmetal / (mmetal•ΔTmetal)

• Cmetal = Qmetal / (mmetal•ΔTmetal) = (313.5 J)/[(12.9 g)•(60.6°C)]

• Solution:• Cmetal = 0.40103 J/g/°C

• Cmetal = 0.40 J/g/°C (rounded to two significant digits)

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 = m•ΔHfusion

• Steps:• Q = m•ΔHfusion = (48.2 g)•(333 J/g)

• Solution:• Q = 16050.6 J• Q = 1.61 x 104 J = 16.1 kJ

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.

EXAMPLE 4:

• Givens:• Given Info about Ice: • m = 50.0 g• ΔHfusion = 333 J/g

• Given Info about Liquid Water: • C = 4.18 J/g/°C • Tinitial = 26.5°C

• Tfinal = 0.0°CΔT = -26.5°C (Tfinal - Tinitial )

• Equation:• The energy gained by the ice is equal to the energy lost from the water. • Qice = -Qliquid 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:

EXAMPLE 4:

• Steps:• Qice = m•ΔHfusion = (50.0 g)•(333 J/g)

• Qice = 16650 J

• Now we can set the right side of the equation equal to m•C•ΔT 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 = -Qliquid water

• 16650 J = -mliquid water•Cliquid water•ΔTliquid water

• 16650 J = -mliquid water•(4.18 J/g/°C)•(-26.5°C)

• 16650 J = -mliquid water•(-110.77 J/°C)

• Solution:• mliquid water = -(16650 J)/(-110.77 J/°C)

mliquid water = 150.311 gmliquid water = 1.50x102 g (rounded to three significant digits)