Specific Heat Capacity. Imagine… A hot day in Arizona…in your back yard is a metal barbeque and a glass of water. Would you want to stick your hand in.

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Presentation transcript:

Specific Heat Capacity

Imagine… A hot day in Arizona…in your back yard is a metal barbeque and a glass of water. Would you want to stick your hand in the water or touch the bbq? You would want to touch the water because different substances have different abilities to absorb heat!!

Specific Heat Capacity The quantity of heat required to raise the temperature of 1 gram of a substance by one degree Celsius (or Kelvin) The quantity of heat required to raise the temperature of 1 gram of a substance by one degree Celsius (or Kelvin) The unit is J/g( o C) The unit is J/g( o C)

Some Specific Heat Capacities Substance Specific heat capacity in J/g( o C) Water4.18 Copper0.385 Iron0.449

Equation: q = mC T q = heat transferred (in joules) q = heat transferred (in joules) m = mass (in grams) m = mass (in grams) C = specific heat capacity C = specific heat capacity T = change in temperature T = change in temperature T = T final - T initial T = T final - T initial

Practice Problem What is the specific heat of silver if the temperature of a 15.4 g sample of silver is increased from 20.0 o C to 31.2 o C when 40.5 J of heat is added?

Givens: m = 15.4 g T i = 20.0 o C T f = 31.2 o C q = 40.5 J q = mC∆T q = mC∆T40.5=15.4(C)( )40.5=15.4(C)(11.2)40.5=172.48(C) C = J/g( o C) C = J/g( o C)

Practice Problem What is the final temp of silver if the temperature of a 5.8 g sample of silver starts out at 30.0 o C and 40.5 J of heat is added? The specific heat of silver is.235 J/g( o C).

Givens: m = 5.8 g T i = 30.0 o C q = 40.5 J C = T f = ??? q = mC∆T q = mC∆T 40.5=5.8(0.235)(T f -30.0) 40.5=1.363(T f -30.0) 40.5=1.363T f – =1.363Tf T f = T f = T f = 60. o C T f = 60. o C

What quantity of heat is required to raise the temperature of 100 mL of water from 45.6  C to 52.8  C? The specific heat of water is J/g(  C) and water has a density of 1.00 grams/mL. What quantity of heat is required to raise the temperature of 100 mL of water from 45.6  C to 52.8  C? The specific heat of water is J/g(  C) and water has a density of 1.00 grams/mL. Practice Problem

Givens: q = ? V = 100. mL T i = 45.6 o C T f = 52.8 o C C = d water =1.00 g/mL 100 mL = 100 g q = mC∆T q=100(4.184)( )q= q=3010 J

Heat of Fusion & Heat of Vaporization Specific heat works great for the part of the heating and cooling process where temperature is changing but what do we do to calculate Q when we are in the process of a phase change? Specific heat works great for the part of the heating and cooling process where temperature is changing but what do we do to calculate Q when we are in the process of a phase change? Answer: We use heat of fusion or heat of vaporization! Answer: We use heat of fusion or heat of vaporization!

Heat of Vaporization The amount of energy needed to vaporize (l  g) a specific amount of a liquid at constant temperature The amount of energy needed to vaporize (l  g) a specific amount of a liquid at constant temperature Unit is usually J/g Unit is usually J/g Q = m(H v )

Example Problem A 115 g sample of liquid is boiled over a 10-minute period. As the liquid boils, the temperature remains constant with 2.27 x 10 5 J of heat is absorbed. At the end of the 10-minute period, all of the liquid has been boiled away. What was the heat of vaporization of the liquid? A 115 g sample of liquid is boiled over a 10-minute period. As the liquid boils, the temperature remains constant with 2.27 x 10 5 J of heat is absorbed. At the end of the 10-minute period, all of the liquid has been boiled away. What was the heat of vaporization of the liquid?

Givens Q = 2.27 x 10 5 J m = 115 g Q = m(H v ) 2.27 x 10 5 = 115(H v ) H v = J/g = 1970 J/g

Heat of Fusion The amount of heat needed to change a specific amount of a solid to a liquid at constant temperature. The amount of heat needed to change a specific amount of a solid to a liquid at constant temperature. Unit is usually J/g Unit is usually J/g Q = m(H f )

Example Problem The heat of fusion of ice at 0 o C is 3.4 x 10 2 J/g. How much heat is needed to change 75g of ice at 0 o C to liquid water at the same temperature?

Givens H f = 3.4 x 10 2 J/g m = 75 g Q = m(H f ) Q = 75(3.4 x 10 2 ) Q = 2.55 x 10 4 J = 2.6 x 10 4 J

Q Total When using multiple equations to solve for the energy change of a system, Q T is used. When using multiple equations to solve for the energy change of a system, Q T is used. Q T = Q 1 + Q 2 + etc.

Practice Problem Water that has a mass of 283g is at 23.5 o C and is placed in a freezer that has a temperature of o C. How much energy must be removed from the water to reach this temperature? Water that has a mass of 283g is at 23.5 o C and is placed in a freezer that has a temperature of o C. How much energy must be removed from the water to reach this temperature?