MEASUREMENT OF HEAT ENERGY
Measurement Of Heat Energy:
Heat energy is the
energy that is transferred from one place another as a result of
temperature difference. Heat energy always flow from a region of high
temperature to the region of lower temperature. Heat energy is called thermal
energy. It is measured in joule.
Heat Capacity:
Heat capacity is the quantity of heat energy
that is require to raise the temperature of a given mass of an object by one
Kelvin. The unit of heat / thermal capacity is joule per Kelvin ( J/K or JK -1
)
Formula Of Heat Capacity, H:
The formula for calculating heat capacity is as
stated .
Heat Capacity Cp = mass (m) x Specific heat
capacity (c) = m x c.
Worked Examples:
I. Calculate
the heat capacity of a mass 258Kg if its specific heat capacity is 900JKg-1K-1.
Solution:
Mass = 258kg, specific heat capacity, c = 900Jkg-1K;
Cp = m x c =258 x 900. ➡ Cp = 232200JK-1
II. The
thermal capacity of an object is 585 JK-1. Calculate the mass of the
object if its specific heat capacity is 390 kg-1K-1
Solution:
Cp =585 JK-1, c= 390 kg-1K-1;
Cp = m x c ; m =
III. Determine
the specific heat capacity of a substance whose mass is 200 grams and of
thermal capacity 250 JK-1
Solution:
M = 200g = 200÷1000 = 0.2 Kg; Cp = 250 JK-1; Cp = m x c ; c =
Specific Heat Capacity:
Specific heat capacity of a substance is the
quantity of heat that is required to raise the temperature of 1 kg mass by 1°C
or 1Kelvin.
The quantity of heat that is required to
raise the temperature of a substance by 1 K is directly proportional to
the mass of the substance, the temperature change the substance .
Mathematically, Quantity of
heat H
H is quantity of heat measured in Joule, m is
mass of substance measured in kg, ∆T is
temperature measured in °C or K.
The unit of specific heat capacity is Joule per
Kilogram per Kelvin ( JK-1K-1 )
Worked Examples:
I. What is the
amount of heat that is required to raise the temperature of 350 g of aluminium
cone from 30°C to 68°C if its specific heat capacity is 900 JKg-1 K-1.
Solution:
M = 350 g = 350/1000 = 0.35 kg, s.h.c = 900 JK-1
K-1, T1 = 30°C,T2 = 68°C. ∆T = T1 -
T2 = 68 – 30 = 38°C
Q =
mc∆T; Q = 0.35 x 900 x 38 = 11970 Joules
II. What is
meant by the statement that the specific heat capacity of water is 4200JKg-1K-1?
Calculate the temperature change when 1000 J of
heat is supplied to 100g of water.
Solution:
Mass = 100g = 100/1000 = 0.1 kg, c =
4200JK-1K-1,Q = 1000; Q = m x c x
∆T; 1000
= 0.1 x 4200 x ∆T; ∆T=
Methods Of Determining The Specific Heat
Capacity Of a Substance:
Different methods can be used to determine the
specific heat capacity of a substance. The specific heat capacity of a
substance can be determined by he following methods:
1. Electrical method 2. Method of mixtures
Determination Of Specific Heat Capacity Of a
Solid By Electrical Method:
In electrical method of determining the specific
heat capacity of a solid substance, electric heater is used to provide
the heat required for the experiment.
Aim: to determine the specific heat capacity of
a solid.
Apparatus:
solid whose specific heat capacity is o be
determined, heater, voltage source, thermometer, calorimeter.
Setup Diagram:
Procedures:
Get a solid whose specific heat capacity you
want to determine and that fits into the calorimeter. Bore two holes in the solid. Weigh and record the mass
of the solid. Insert the thermometer and the heater into the holes and add
little oil to help establish good thermal contact between the block and the
thermometer and heater. Read and record
the initial temperature of the solid and calorimeter. Switch on the electrical
heater so that current flows for some times until the temperature rise is about
15°C. Use a stop watch to measure the time for which current flows. Read and
record the final temperature of the solid and thermometer.
Data From the Procedures:
Mass of solid = Ms; Mass of calorimeter = Mc ; Initial
temperature of solid and calorimeter = T1
Final temperature of solid and calorimeter = T2
; Voltage
applied across heater = V Current that flow = I
Time for which current flow = t
Theory of calculation:
Heat supplied by heater = heat gained by solid +
heat gained by calorimeter.
IVt = Ms
x Cs x ∆T + Mc x Cc x ∆T
Ms = mass of solid. Cs = specific heat capacity.
Mc = mass of calorimeter. Cc = specific heat capacity of calorimeter. ∆T
= temperature change. I = current.
V = voltage. t = time.
Precautions:
1. Make
sure that the calorimeter is lagged to prevent heat lose. 2. Take the
reading when the mercury thread is steady.
3. Avoid
error due to parallax when taking the reading.
Worked Examples:
1. An
electric heater, rated 20V, 40 W, fitted into a metal block supplied heat to
the block of mass 2.25 kg and specific heat capacity of 460JKg-1K-1.
Calculate the temperature rise in the block if the current flow for 10 minutes.
Solution:
Voltage = 20 V, power = IV = 40 W, mass
=2.25kg, C = 460 JK-1K-1 , t = 10 minutes = 10 x 60 = 600 seconds.
IVt = Ms x Cs x ∆T + Mc x Cc x ∆T
IVt = Ms x Cs x ∆T; P x t = Ms x Cs x ∆T; 40 x 600
= 2.25 x 640 x ∆T; ∆T=
Determination
Of Specific Heat Capacity Of a Liquid By Electrical Method:
In electrical method of determining the specific heat capacity of a
liquid substance, electric heater is used to heat the liquid during the
experiment.
Aim: to determine the specific heat capacity of
a liquid.
Apparatus: liquid whose specific heat capacity
is o be determined, heater, voltage source, thermometer, calorimeter.
Setup Diagram:
Procedures:
Weigh and record the mass of empty calorimeter.
Fill the calorimeter with a liquid whose specific heat capacity you want to
determine. Weigh and record the mass of the calorimeter and the liquid. Insert
the thermometer and the heater into the liquid in the calorimeter.
Read and record the initial temperature of the
liquid and calorimeter. Switch on the electrical heater so that current flows
for some times until the temperature rise is about 15°C. Use a stop watch to
measure the time for which current flows. Stir the liquid for equal temperature.
Read and record the final temperature of the liquid and the calorimeter.
Data From the Procedures:
Mass of empty calorimeter = Mc Mass
of liquid and calorimeter = M LC Mass of liquid = MLC
- MC = ML
Specific heat capacity of liquid = CL
Specific heat capacity of calorimeter = CC Initial
temperature of liquid and calorimeter = T1
Final temperature of liquid and calorimeter = T2
Voltage
applied across heater = V
Current that flow = I
Time for which
current flow = t
Theory of calculation:
Heat supplied by heater = heat gained by liquid
+ heat gained by calorimeter.
IVt = ML x CL x ∆T + Mc x Cc x ∆T
Make CL the of formula: IVt
- Mc x Cc x ∆T = ML x CL x ∆T; CL=
ML = mass of liquid. Cs = specific heat
capacity. Mc = mass of calorimeter. Cc
= specific heat capacity of calorimeter.
∆T = temperature
change. I = current. V = voltage. t = time.
Precautions:
1. Stir the liquid continuously for equal
temperature. 2.Take the reading of the thermometer when the mercury
thread is steady.
3. Avoid
error due to parallax when taking the reading of the thermometer. 4. Make sure the calorimeter is properly lagged.
5. Do not use large
current or voltage so as not o damage the appliance.
Worked Examples:
350g of water is heated so that its temperature
rises from 30°C to 67°C in 35 minutes. Calculate the quantity of heat supplied
and the heat supplied per minute. ( s.h.c. of water = 4200 JK-1K -1 )
Solution:
Mass = 350g = 350/1000 = 0.35kg, T1=
30°C, T2 = 67°C, Cw = 4200JK-1 K-1 , ∆T = T2 –
T1 = 67 – 30 = 37°C.
Quantity of heat Q = ML x CL x ∆T; Q = 0.35
x 4200 x 37 = 54390 Joules.
Quantity of heat per minute = total quantity of
heat / total time = 54390/7 x 60 = 129.5 Joules. 7.
Assignment
1. A tap supplies water at 26°C while another supplies at 83°C. If a man wishes to bath with water at 40°C, what is the ratio to mass of hot water to that of cold water?
2. When two objects p and q are supplied with same quantity of heat, the temperature change in p is twice that of q. If the masses of p and q are the same, calculate the ratio of the specific heat capacities of q to p.
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