Heat and temperature are not the same thing.
It takes more heat to change the temperature of a full saucepan of water than it does to change the temoerature of an empty pan by the same amount. Heat flows from hot to cold objects: it is the transfer of energy due to a temperature difference. "Cold" does
not flow from cold objects to hot objects!
The temperature change brought about by supplying an amount of heat depends on:
- the substance being heated up (it takes more heat to warm up the water inside the saucepan than the metal of the saucepan itself
- the amount of the substance (it takes more heat to warm up a full saucepan than a half empty one)
Copper has a
low heat capacity meaning that a small amount of heat produces a large temperature change. Water has a
high heat capacity meaning that a large amount of heat is required to warm it up.
We use two common measures of amount of stuff in science: mass and number of moles. Depending on which we are using, there are therefore two different heat capacity values for each substance:
- If the amount of material is known in grams, we use the specific heat capacity. This has the symbol c and units J g-1 K-1.
- If the amount of material is known in moles, we use the molar heat capacity. This has the symbol C and units J mol-1 K-1.
Substance |
Phase |
c (J g-1 K-1) |
C (J mol-1 K-1) |
Aluminium |
solid |
0.897 |
24.2 |
Copper |
solid |
0.385 |
24.47 |
Diamond |
solid |
0.5091 |
6.115 |
Ethanol |
liquid |
2.44 |
112 |
Glass |
solid |
0.84 |
|
Gold |
solid |
0.129 |
25.42 |
Granite |
solid |
0.790 |
|
Graphite |
solid |
0.710 |
8.53 |
Iron |
solid |
0.450 |
25.1 |
Lead |
solid |
0.127 |
26.4 |
Silver |
solid |
0.233 |
24.9 |
Water at 25 °C |
liquid |
4.1813 |
75.327 |
The heat,
q is thus related to the temperature change, Δ
T, by:
For a mass m: | | For a number of moles n: |
q = m × c × ΔT |
|
q = n × C × ΔT |
ΔT = q / (m × c) |
|
ΔT = q / (n × C) |
Points to note:
- ΔT is the temperature difference between the initial temperature, Ti, and the final temperature, Tf such that:
ΔT = ΔTf - ΔTi
- If the temperature increases, Tf > Ti and so ΔT > 0 and so q > 0. Increasing the heat, increases the temperature and vice versa.
- If the temperature decreases, Tf < Ti and so ΔT < 0 and so q < 0. Decreasing the heat, decreases the temperature and vice versa.
- If Tf and Ti are given in °C, there is no need to convert each to Kelvin as the difference between them is the same on the centigrade and Kelvin scales.
- Whenever you calculate a property from a measurement, the signficant figures must be handled correctly, read the iChem online tutorial (identical to the E9 week 4 pre-laboratory information).
What is the final temperature of 12 g of water, initially at a temperature of 25.00 °C, if 125 J of heat is added to it? (The specific heat capacity of water is 4.1813 J g-1 K-1.)
As we are given the mass of water and want to calculate the temperature chsnge:
ΔT = q / (m × c) = (125 J) / (12 g × 4.1813 J g-1 K-1) = 2.5 K or 2.5 °C
The mass is only known to 2 significant figures so the temperature change is also known only to 2 significant figures. Note that as heat is added, q is positive leading to ΔT being positive too.
The initial temperature is 25.00 °C and the temperature increase is positive so the final temperature is (25.00 + 2.5) °C = 27.5 °C
What is the heat change for a 25 g block of aluminium, initially at a temperature of 225 °C, plunged into a large bath of water at 26 °C? (The specific heat capacity of aluminium is 0.897 J g-1 K-1.)
The initial temperature is 225 °C and the final temperature is 26 °C. Hence the temperature change is:
ΔT = Tf - Ti = (26 - 225 °C) = -199 °C or -199 K.
As we are given the mass of water and want to calculate the heat chsnge:
q = m × c × ΔT = 25 g × 0.897 J g-1 K-1 × -199 K = -4500 J
The mass is only known to 2 significant figures so the heat change is also known only to 2 significant figures. Note that the aluminium cools and heat is lost, ΔT is negative leading to q being negative too.
1500 J of heat are added to a block of copper resulting in its temperature rising by 250 ° C. How many moles of copper are in the block? (The molar heat capacity of copper is 24.47 J mol-1 K-1.)
We are given the heat and temperature changes and want to calculate the amount of copper present. As the number of moles is requested, the molar heat capacity is used:
n = q / (C × ΔT) = (1500 J) / (24.47 J mol-1 K-1 ×250 K) = 0.25 mol
The heat and temperature changes are both only known to 2 significant figures (the trailing zeros are not significant) so the number of moles is also known only to 2 significant figures
