Heat is not the same thing as temperature , even though in common usage these concepts are often used interchangeable. Heat is the energy transferred from one object to another due to a difference in their temperature. Heat, therefore, has units of energy (joules, J). An object at a higher temperature will transfer energy to one at a lower temperature until they reach thermal equilibrium – until they are at the same temperature.
The amount of heat gained (to raise the temperature) or lost (to lower the temperature) by an object can be quantified with the following equations:
(1) 𝑞 = 𝑚𝑐∆𝑇 or (2) 𝑞 = 𝑛𝐶∆𝑇 where q is the heat change (in J), m is the mass (in g), n is the number of moles (in mol), c is the specific heat capacity (in J g -1 K -1 ) and C is the molar heat capacity (J mol -1 K -1 ).
The change in temperature, Δ T , is always:
Δ
T = T final
– T initial
Hence, if the temperature increases, Δ T is positive and, if the temperature decreases, Δ T is negative.
The two equations (1) and (2) will give the same value for q as long as the specific heat capacity is used when you know the mass and the molar heat capacity is used when you know the number of moles.
The specific and molar heat capacities are a measure of how much energy is needed to raise the temperature of 1 g or 1 mol, respectively, of an object by 1 K. Every object has a different heat capacity: some substances, like metals, are easier to heat than others, like rocks.
1.
Under what circumstances would you get a negative value for the heat, q ?
2.
Provide an equation for converting between the specific heat capacity, c , and the molar heat capacity, C .
( Hint : use the relationship between the number of moles and the mass of a substance).
3.
If the temperature of a substance increases from 25 °C to 35 °C, what is Δ
T (in Kelvin)?
4.
Given your answer to Q3, explain to your group whether it is necessary to convert temperatures to
Kelvin when working out Δ
T .
5.
The specific heat capacity of olive oil is 2.0 J g -1 K -1 . How much energy has to be transferred to 2.0 g of olive oil in a saucepan to heat it from room temperature to 130 °C? Assume that the room is at 25 °C.
6.
The specific heat capacity of water is 4.18 J g
-1
K
-1
. Is it easier or harder to heat water or olive oil?
7.
The molar heat capacity of gold is 25.413 J mol
-1
K
-1
. A necklace that weighs 1.2 g requires 0.426 J of energy to heat by 2.00 K. Is the necklace pure gold? ( Hint : you will first need to either convert the molar heat capacity to the specific heat capacity using the equation you worked out in Q2 or convert the mass into the moles).
Research Project on Global Warming Assignment - deadline is the end of week 11

8.
A calorimeter containing 300.0 mL of water at 25 °C was calibrated as follows. A 1000.0 W heating coil was run for 10.0 s, after which time the temperature had increased by 7.5 °C. Calculate the heat capacity of the empty calorimeter. The specific heat of water is 4.184 J K
–1
g
–1
.
9. If 15.0 g of sodium nitrite was dissolved into this calorimeter, and the temperature of the solution was found to decrease by 2.6

C, calculate the enthalpy of solution of sodium nitrite.
10. Calcium chloride (1.14 g) is completely dissolved in 100.0 mL of water at 27.00 °C in a ‘coffee cup’ calorimeter. The temperature of the water after dissolution is 28.97 °C. Calculate the standard enthalpy of solution of CaCl
2
(in kJ mol heat capacity is 4.184 J K
–1
g
–1
–1 ). The density of water at 27.0 °C is 0.997 g mL
. Ignore the heat capacity of the CaCl
2
.
–1
and its
Research Project on Global Warming Assignment - deadline is the end of week 11

11. A new process has been developed for converting cellulose from corn waste into the biofuel butanol,
C
4
H
9
OH. A bomb calorimeter with a heat capacity of 3250 J K
–1
was used to determine the calorific value by burning 5.0 g of butanol in excess oxygen.
(a) Write a balanced reaction for the combustion of butanol in oxygen.
(b) Calculate the heat released from this combustion if the temperature of the calorimeter increased from 23.0 to 78.6 °C during the test.
(c) Use this value to determine the calorific value and molar enthalpy of combustion of butanol.
( Hint : the calorific value is the energy content per gram.)
Planets are in thermal equilibrium. The energy they absorb from the star they orbit is in equilibrium with the amount of energy they emit.
The Stefan-Boltzmann law states that the total energy radiated per unit surface area of a black body per unit time is directly proportional to the fourth power of the black body’s temperature. If a star, such as the sun, is considered to be a black body, then this law gives:
P out
= 4π r
2
×
σ
× T
4
2. where
σ
is the Stefan-Boltzmann constant which is equal to 5.67 × 10
-8
J m
-2 s
-1
K
-4
1. What does each symbol represent and what are the associated SI units?
If the radius of the Sun is 696 × 10 6
m and its surface temperature is 5780 K, what is P out
?
Research Project on Global Warming Assignment - deadline is the end of week 11

This is the energy released by the Sun at its surface. The amount of energy received by a planet which orbits at a distance d from the Sun is: 𝑞 𝑜𝑢𝑡
I in
=
4𝜋𝑑 2
If the planet has a radius r , then the total amount of solar energy absorbed across its cross-section is
P in
= I in
× π r
2
3.
If the Earth is 149.598 × 10 9
m from the Sun and has a radius of 6378 × 10
3
m, what is P in
for the
Earth?
The Earth can also be treated as a black body then it emits energy such that P out
= 4π r
2
×
σ
× T
4
. If all of the energy from the Sun is completely absorbed, then P in
, is equal to P out
.
4. Using your answer to Q3, calculate the temperature of the Earth.
5. Actually, the Earth reflects about 28% of the sunlight straight back into space. Recalculate your answer for Q3.
6. Using the value of P in
from Q5, recalculate the temperature of the Earth.
7. If half of the polar ice caps were to melt, what would happen qualitatively to the percentage reflection by the Earth? Predict the effect on the temperature of the planet and the effect on the remaining ice cap.
8. The average temperature of the Earth is about 290 K. This is because not all of the radiation emitted by the surface escapes into space. What might cause this?
Research Project on Global Warming Assignment - deadline is the end of week 11