1. Calorimetry College Chemistry

2. 6. Specific Heat a. Some things heat up or cool down faster than others. Land heats up and cools down faster than water

3. b. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K). 1) c water = 4184 J / kg C 2) c sand = 664 J / kg C This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

4. How to calculate changes in thermal energy Q = m x  T x c Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) c = specific heat of substance

5. c. A calorimeter is used to help measure the specific heat of a substance. First, mass and temperature of water are measured Then heated sample is put inside and heat flows into water  T is measured for water to help get its heat gain This gives the heat lost by the substance

6. Calorimetry Calorimetry means “measuring heat” In practice, it is a technique used to measure specific heat Technique involves: Raising temperature of object(s) to some value Place object(s) in vessel containing cold water of known mass and temperature Measure temperature of object(s) + water after equilibrium is reached A calorimeter is a vessel providing good insulation that allows a thermal equilibrium to be achieved between substances without any energy loss to the environment (styrofoam cup or thermos with lid) Conservation of energy requires that: ( Q > 0 ( < 0 ) when energy is gained (lost))

7. Example Problem A 0.400 kg block of iron is heated from 295 K to 325 K. How much heat had to be transferred to the iron? Q = mc  T Q = (0.400 kg)(450 J/kg*K)(325-295 K) Q = 5.4 x 10 3 J Mass can be in grams OR kilograms  MUST match with specific heat! Specific heat is a constant  it will be given to you OR you will solve for it

8. Example 1. How much energy is required to heat 955g from 20ºC to 100ºC? Specific heat= 4.184J/K*g Q=mcΔT Q=(955g)(4.184J/K*g)(80) Q=319.7kJ You do NOT need to convert to Kelvin because it’s a difference in temperature, if it asked to find a temp, you would have to

9. Specific Heat vs. Heat Capacity Specific heat – (c) amount of heat required to raise the temperature of 1 gram by 1 degree Celsius Heat capacity – (C) amount of heat required to raise the temperature of a specific quantity of substance by 1 degree Celsius We often care about the heat capacity of a calorimeter  it lets us determine how much heat is leaving the calorimeter C = mc Or we can substitute heat capacity in q=ms  T q = C  T

10. Constant-Volume Calorimetry Heat of combustion usually measured in a constant volume calorimeter Place a known mass in a steel container filled with oxygen and a known amount of water Sample ignited electrically and the heat produced can be accurately calculated by computing the rise in temperature of the water q cal = C cal  t

11. Example 6.6 A quantity of 1.435 g of naphthalene (C 10 H 8 ) was burned in a constant-volume bomb calorimeter. The temperature rose from 20.28°C to 25.95°C. If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion on a molar basis (molar heat of combustion). q = C  t = (10.17 kJ/°C)(25.95 – 20.28°C) = 57.6 kJ Per mole: -57.66 kJ 128.2 g C 10 H 8 1.435 g C 10 H 8 1 mol C10H8 -5.15 x 10 3 kJ/mol

12. Constant Pressure Calorimetry Uses a Styrofoam cup  measures heat of solution and heat of dilution Since pressure is constant, q =  H

13. Another Calorimetry Problem A 25g piece of iron (s=.4494 J/g*C) has been heated to 95C and dropped into a 100mL of 15C water (s=4.184 J/g*C). What is the final temperature of the system? -q (water) =q (Fe) OR Σq water + q Fe - q (water) = mcΔT=(100)(4.184)(T f -15) = q (Fe) = mcΔT=(25)(0.4494)(T f -95) Solve for T f = 17.1 C