Heat, Temperature, and Energy Transfer
Heat and temperature
Heat
, q, is thermal energy transferred
(between two systems that are
different in temperature) or from
hotter system to cooler system that
are in contact.
•
Heat and temperature are two different but
closely related concepts. Note that they have
different units: temperature typically has units
of degrees or Kelvin (K), and heat has units of
energy, Joules (J). Temperature is a measure of
the average kinetic energy of the atoms or
molecules in the system.
•
T (C˚) + 273.15 = T (K)
Relationship between heat and
temperature
Heat capacity
SPECIFIC HEAT CAPACITY
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Lower specific heat capacity heats
up and cools down quickly because
it takes less energy to change its
temperature one degree.
Higher specific heat capacity heats up
and cools down slowly because it takes
more energy to change its temperature
one degree.
•
Smaller mass, heats up faster and cool down
faster. Larger mass heats up slower and cool
down slower.
•
We can calculate the heat released or
absorbed using the specific heat capacity C
equation:
•
q = m × C × ΔT
•
M is the mass of the substance
•
ΔT is the change in temperature
•
The unit of heat is joul J or KJ
For example, if we have a cup of hot
coffee and cold tea. Water molecules
in a cup of hot coffee have a higher
average kinetic energy than water
molecules in a cup of cold tea, which
means that they are moving at a
higher velocity.
•
When a system absorbs or loses heat, the
average kinetic energy of the molecules will
change. Thus, heat transfer results in change in
the system's temperature
as long as the system is
not undergoing a phase change.
•
Example:
•
we have 250 mL of hot tea which we would like
to cool down before we drink it. The tea is
currently at 370 K, and we'd like to cool it down
to 350 K. How much thermal energy has to be
transferred from the tea to the surroundings to
cool the tea?
•
We are going to assume that the tea is mostly
water, so we can use the density and heat
capacity of water in our calculations. The specific
heat capacity of water is 4.18 J/g
⋅
K, and the
density of water is 1.00 g/ml. We can calculate
the energy transferred in the process of cooling
the tea using the following steps:
•
1. Calculate the mass of the substance
•
We can calculate the mass of the tea/water using
the volume and density of water:
•
m=250 mL×1.00 g/mL=250g
•
2. Calculate the change in temperature
Δ
T
•
We can calculate the change in temperature,
Δ
T,
from the initial and final temperatures:
•
Δ
T= T
final
– T
initial
•
=350K−370K
•
=−20K
3. Solve for q
•
Now we can solve for the heat transferred from
the hot tea using the equation for heat:
•
Q= m x c x ∆T
•
= 250 g X 4.18 J/ g.K X (-20) K
•
= -21000J
•
Thus, we calculated that the tea will
transfer 21000 J of energy to the surroundings
when it cools down from 370 K to 350 K.
Delve into the concepts of heat and temperature, exploring their relationship and implications in energy transfer. Discover the significance of heat capacity, specific heat capacity, and how mass influences the heating and cooling rate of substances. Learn how to calculate heat using specific heat capacity equations and gain insights into scenarios where heat influences the behavior of molecules.
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Presentation Transcript
Heat and temperature Heat, q, is thermal energy transferred (between two systems that are different in temperature) or from hotter system to cooler system that are in contact.
Relationship between heat and temperature Heat and temperature are two different but closely related concepts. Note that they have different units: temperature typically has units of degrees or Kelvin (K), and heat has units of energy, Joules (J). Temperature is a measure of the average kinetic energy of the atoms or molecules in the system. T (C ) + 273.15 = T (K)
Heat capacity Heat capacity is the amount of heat to be supplied to material to produce a unit change in its temperature. C = ? ? The unit of heat capacity is joule per kelvin.
SPECIFIC HEAT CAPACITY the specific heat capacity of a substance is the heat capacity of substance divided by the mass of the sample. a sample of the
Lower specific heat capacity heats up and cools down quickly because it takes less energy to change its temperature one degree. Higher specific heat capacity heats up and cools down slowly because it takes more energy to change its temperature one degree.
Smaller mass, heats up faster and cool down faster. Larger mass heats up slower and cool down slower.
We can calculate the heat released or absorbed using the specific heat capacity C equation: q = m C T M is the mass of the substance T is the change in temperature The unit of heat is joul J or KJ
For example, if we have a cup of hot coffee and cold tea. Water molecules in a cup of hot coffee have a higher average kinetic energy than water molecules in a cup of cold tea, which means that they are moving at a higher velocity.
When a system absorbs or loses heat, the average kinetic energy of the molecules will change. Thus, heat transfer results in change in the system's temperature as long as the system is not undergoing a phase change. Example: we have 250 mL of hot tea which we would like to cool down before we drink it. The tea is currently at 370 K, and we'd like to cool it down to 350 K. How much thermal energy has to be transferred from the tea to the surroundings to cool the tea?
We are going to assume that the tea is mostly water, so we can use the density and heat capacity of water in our calculations. The specific heat capacity of water is 4.18 J/ g K, and the density of water is 1.00 g/ml. We can calculate the energy transferred in the process of cooling the tea using the following steps: 1. Calculate the mass of the substance We can calculate the mass of the tea/water using the volume and density of water:
m=250 mL1.00 g/mL=250g 2. Calculate the change in temperature T We can calculate the change in temperature, T, from the initial and final temperatures: T= Tfinal Tinitial =350K 370K = 20K 3. Solve for q Now we can solve for the heat transferred from the hot tea using the equation for heat:
Q= m x c x T = 250 g X 4.18 J/ g.K X (-20) K = -21000J Thus, we calculated that the tea will transfer 21000 J of energy to the surroundings when it cools down from 370 K to 350 K.
