Welcome to the MATH

Specific Heat Capacity- The amount of heat it takes to raise the temperature of 1 gram of a substance by 1OC. Abbreviated “C” Water has a HUGE value, compared to other chemicals.

For water, C = 4.18 J/(gOC) This means, for water: It takes a long time to heat up, and It takes a long time to cool off. Water is used as a coolant!

To calculate, use the formula: q = m x C x ΔT Heat is abbreviated as “q” and the units are Joules. m = mass in grams ΔT = change in temperature, calculated as Tfinal – Tinitial C = specific heat units are in J/gOC

Heating Curve

The flat areas of the graph represent phase changes. The formula for the heat in a phase change involves a concept called enthalpy.

The sign Heat values can have a + value or – value. How can I tell?

The sign You must consider the system and the surroundings. The system is what you are studying, the surroundings is the rest of the universe.

The sign The System

The System The sign Heat will be +, if it is moving into the system. (Temperature is rising) The System

The System The sign Heat will be -, if it is moving out of the system. (Temperature is dropping) The System

Heat in phase changes q = m x ΔH q = heat (in Joules) m = mass (in grams) ΔH = enthalpy (in J/g). This is a constant to be looked up in a chart.

Heat in phase changes ΔHfus = enthalpy of fusion, used when melting. ΔHvap = enthalpy of vaporization, used when boiling.

Calculating Heat How much heat is needed to raise the temperature of 55.75 grams of liquid water from 13.5 oC to 75.0 oC? The specific heat of water is 4.18 J/g oC.

Calculating Heat How much heat is needed to melt 150.0 grams of ice? ΔHfus = 334 J/g

Review How much heat is needed to raise the temperature of 75.0 grams of liquid water from 25.0 oC to 100.0 oC? How much heat is needed to boil it?

Calorimetry Involves using an insulated container and water to find the heat lost by a hot object. Uses the conservation of energy.

Calorimetry The System

Calorimetry Heatloss = Heatgain So the heat lost by a hot object placed in water is equal to the heat gained by the water.

Calorimetry Heatloss = Heatgain m x C x ΔT = m x C x ΔT hot object water Hot side gets a negative sign!! Final temperature is the same for both objects.

Let’s try one If a piece of iron with a mass of 23.5 grams and a temp. of 455 oC is dropped into 150.0 grams of water at 25.0 oC, what is the final temperature of the system? C of iron = 0.450 J/g oC C of water = 4.18 J/g oC

You try one If a piece of iron with a mass of 46.5 grams and a temp. of 100 oC is dropped into 150.0 grams of water at 25.0 oC, what is the final temperature of the system? C of iron = 0.450 J/g oC C of water = 4.18 J/g oC

You try one # 2 A piece of an unknown metal with a mass of 75.5 grams and a temp. of 255oC is dropped into 150.0 grams of water at 25.0 oC. If the final temperature of the system is 27.9oC, find the C of the metal. C of water = 4.18 J/g oC

Review If a piece of lead with a mass of 93.7 grams and a temp. of 250.0 oC is dropped into 98.75 grams of water at 22.4 oC, what is the final temperature of the system? C of Pb = 0.128 J/g oC C of water = 4.18 J/g oC

Review A chunk of iron is heated from 25.0oC to 1536oC. If the mass is 5.75 grams, how much heat is transferred? How much heat is needed to melt our 5.75 gram piece of iron? C of iron = 0.450 J/g oC ΔHfus for iron = 13.8 kJ/mol

Review How much heat is needed to melt our 5.75 gram piece of iron? ΔHfus for iron = 13.8 kJ/mol

Review A piece of Aluminum metal with a mass of 25.0 grams and a temp. of 475oC is dropped into 75.0 grams of water at 22.5 oC. What is the final temperature of the system? C of water = 4.18 J/g oC C of aluminum = 0.9025 J/g oC