What is Calorimetry?

Salvador Dali, Persistence of Memory, 1931

Review

1. What is the specific heat of a 10. g sample of a substance that requires 950 J of heat to raise its temp from 18 °C to 42 °C?

2. Mercury (Hg) has a specific heat of 0.14 J/g°C. How much energy does a 22.8 g sample of mercury lose as its temp goes from 33 °C to 16 °C ?

Calorimetry

Calorimeter – an instrument that measures �the temperature change due to the gain or loss of heat.

A calorimeter can be used to determine the specific heat of a substance, or to determine the heat given off or absorbed by a chemical rxn.

In calorimetry, the heat change for

a process is determined by

measuring the heat change in

the water surrounding the process

in an insulated container called

a calorimeter.

Determining the specific heat of an object

using calorimetry:

q_o = - q_w

heat lost by object = - heat gained by water

What is meant by ΔT?

• As you may or may not remember from your math class, Δ means “change”.
• Change always means

final condition – initial condition.

• In this case that means:
• ΔT = T_final - T_initial

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?
2. What metal is it?

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?

m_o x C_o x ΔT_o = - m_w x C_w x ΔT_w

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?

m_o x C_o x ΔT_o = - m_w x C_w x ΔT_w

29.5 g x C_o x (12.00°C – 100.0 °C) =

-(150.0 g x 4.18 J/g°C x (12.00 °C – 10.00 °C)

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?

m_o x C_o x ΔT_o = - m_w x C_w x ΔT_w

29.5 g x C_o x (12.00°C – 100.0 °C) =

-(150.0 g x 4.18 J/g°C x (12.00 °C – 10.00 °C)

​

Simplify and solve for C_o.

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?

m_o x C_o x ΔT_o = - m_w x C_w x ΔT_w

29.5 g x C_o x (12.00°C – 100.0 °C) =

-(150.0 g x 4.18 J/g°C x (12.00 °C – 10.00 °C)

​

Simplify and solve for C_o.

-2596 x C_o = -1254

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

1. What is the specific heat of the metal?

m_o x C_o x ΔT_o = - m_w x C_w x ΔT_w

29.5 g x C_o x (12.00°C – 100.0 °C) =

-(150.0 g x 4.18 J/g°C x (12.00 °C – 10.00 °C)

​

Simplify and solve for C_o.

-2596 x C_o = -1254

C_o = 0.483 J/g°C

Example 1: A 29.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 150.0 g of water at 10.00 °C. The temperature of the water rises to 12.00 °C.

b) What metal is it?

If C_o = 0.483 J/g°C, look on a table of C values and find the metal that matches. In this case, what would it be? (Pick the closest!)

Example 2: A 12.5 g sample of metal with a temperature of 100.0 °C is placed in a calorimeter holding 75.0 g of water at 22.0 °C. The temperature of the water rises to 24.0°C. What is the specific heat of the metal?

​

Answer: 0.66 J/g^oC