Physics Topics can help us understand the behavior of the natural world around us.
What is the Unit of J in Heat?
The widely used unit of heat is calorie (or cal, in CGS system). We know that is a form of energy; so energy units like erg (CGS system) and joule (J, in SI) should somehow be related to the units of heat.
James prescott Joule was the first experimentalist who accurately measured the mechanical energy equivalent to some amount of heat energy i.e., the mechanical equivalent of heat. This measurement determines the amount of mechanical energy expressed in joule (or erg) which is equivalent to heat energy of 1 cal.
Mechanical energy may be used to do work and work produces heat. On the other hand, heat may be converted to work or mechanical energy. This indicates that there is a natural relationship between work and heat. This relation is known as Joule’s law.
Joule’s law:
If some amount of work is entirely converted to heat, then the work done and the heat produced are proportional to each other.
If W = work done and H = heat produced, then from Joule’s law,
W ∝ H or, W = JH ………. (1)
Here, J is a constant. This constant is called the mechanical equivalent of heat or Joule’s equivalent.
Definition of J and its magnitudes in different systems:
In equation (1), if H = 1, then W = J. So mechanical equivalent of heat is defined as the amount of work done to produce a unit amount of heat.
Clearly, the unit and the magnitude of J depends on the unit of heat and work in different systems of units.
CGS system: The unit of H is calorie and the unit of W is erg. As J = \(\frac{W}{H}\), the unit of J is erg ᐧ cal-1 and the measured value of J = 4.2 × 107 erg ᐧ cal-1.
As 1 J = 107 erg, the value of J = 4.2 J ᐧ cal-1.
The units like erg ᐧ cal-1 and J ᐧ cal-1 indicate that the mechanical equivalent of heat J actually denotes the conversion factor between two units. For example, J = 4.2 J ᐧ cal-1.
This means that 1 cal = 4.2 J, i.e., each calorie should be multiplied by 4.2 to express the energy in joule.
In SI: here W and H are expressed in the same unit, which is joule (J). So, no conversion factor is necessary. The mechanical equivalent of heat is J = 1. This means that the concept of J is unnecessary in SI.
Relation of calorie with erg and joule:
1 cal = 4.2 × 107 erg = 4.2 J.
Numerical Examples
Example 1.
800 g lead balls are kept in a 1 m long, vertical tube which is a bad conductor and the tube is then closed at both ends. The tube is suddenly inverted so that the balls fall from one end to the other. The temperature of the lead balls increases by 3.89°C after 50 such inversions. Find out the mechanical equivalent of heat assuming that the lead balls have retained the entire amount of heat produced.
Solution:
Work clone, W= loss of potential energy of the halls
= n × mgh
= 50 × 800 × 980 × 1oo erg (1 m = 100 cm)
Heat produced H = mst = 800 × 0.03 × 3.89 cal
∴ J = \(\frac{W}{H}\) = \(\frac{50 \times 800 \times 980 \times 100}{800 \times 0.03 \times 3.89}\) = 4.2 × 107 erg ᐧ cal-1.
Example 2.
Water drops from a height of 50 m in a waterfall. Find out the rise in temperature of water, if 75% of its energy is converted into heat and absorbed by water. (J = 4.2 × 107 erg ᐧ cal-1; g = 9.8 m ᐧ s-2)
Solution:
Here, W = mgh × \(\frac{75}{100}\) and H = mst
Now, W = JH or, mgh × \(\frac{75}{100}\) = Jmst or, t = \(\frac{m g h \times \frac{75}{100}}{J m s}\)
[s = 1 cal ᐧ g-1 ᐧ °C-1 for water, g = 9.8 m ᐧ s-2 = 980 cm ᐧ s-2, h = 50 m = 50 × 100 cm]
= \(\frac{g h}{J s}\) × \(\frac{75}{100}\) = \(\frac{980 \times 50 \times 100}{4.2 \times 10^7 \times 1}\) × \(\frac{75}{100}\) = 0.0875°C.
Example 3.
The velocity of a 42 kg celestial body reduces from 20 km ᐧ min-1 to 5 km ᐧ min-1 due to its passage through the earth’s atmosphere. Find out the heat produced in calorie. (J = 4.2 × 107 erg ᐧ cal-1
Solution:
Initial velocity of the celestial body
u = 20 km ᐧ min-1 = \(\frac{20 \times 1000}{60}\) m ᐧ s-1;
Final velocity of the celestial body,
v = 5 km ᐧ min-1 = \(\frac{5 \times 1000}{60}\) m ᐧ s
Work done, W = change in kinetic energy
Example 4.
Find out the amount of work done to convert 1oo g ice at 0°C to water at 100 C. (Latent heat of fusion of ice = 80 cal ᐧ g-1; mechanical equivalent of heat = 4.2 J ᐧ cal-1).
Solution:
Total heat supplied,
H = 100 × 80 + 100 × 1 × (100 – 0)
= 8000 + 10000 = 18000 cal
∴ Work done, W = JH = 4.2 × 18000 = 75600 J.
Example 5.
What will be the temperature difference between the top and the bottom of a 400 m high waterfall, assuming that 80% of the heat produced is retained by the water?
Solution:
Work done, W= loss in potential energy
= mg(h – 0) = mgh
∴ Heat produced, H = \(\frac{W}{J}\) = \(\frac{m g h}{J}\)
∴ Amount of heat retained by water = \(\frac{m g h}{I}\) × \(\frac{80}{100}\)
If the increase in temperature of water is t, required heat = mst
According to the question,
mst = \(\frac{m g h}{J}\) × \(\frac{80}{100}\)
or, t = \(\frac{g h}{J s}\) × \(\frac{80}{100}\) = \(\frac{9.8 \times 400}{4.2 \times 1000}\) × \(\frac{80}{100}\) [∵ specific heat of water in S, s = 1000 cal ᐧ kg-1 ᐧ °C-1]
or, t = 0.747°C
Example 6.
Find out the minimum height from which a piece of ice at 0°C should be dropped so that it melts completely due to its impact with the ground. Assume that half of the energy loss due to the fall is responsible for the fusion of ice. (Latent heat of fusion of ice = 80 cal ᐧ g-1, g = 980 cm ᐧ s-2, J = 4.2 J ᐧ cal-1)
Solution:
Let the price of ice of mass m be allowed to fall from height h.
Now, energy loss due to the fall = loss in potential energy = mgh = work done, W
Half of it, i.e., \(\frac{W}{2}\) amount of energy is converted into heat and it is responsible for the fusion of ice.
∴ Heat produced, H = \(\frac{W / 2}{J}\) = \(\frac{m g h}{2 J}\).
As m g of ice melts into water, required latent heat = 80m cal
So, \(\frac{m g h}{2 J}\) = 80m
or, h = \(\frac{80 \times 2 J}{g}\) = \(\frac{80 \times 2 \times 4.2 \times 10^7}{980}\)
[J = 4.2 J ᐧ cal-1 = 4.2 × 107 erg ᐧ cal-1]
= 6.857 × 106 cm = 68.57 km.
Example 7.
A piece of ice at 0°C is dropped to the ground from some height. The piece of ice melts completely due to its impact with the ground. Find the height from which the piece was dropped considering that 60% of its energy is converted into heat. (J = 4.2 J ᐧ cal-1) [HS’ 03]
Solution:
The potential energy of the piece of ice of mass m at height h = mgh, kinetic energy = 0
So, total mechanical energy = mgh
This energy is conserved till it touches the ground. Due to the impact with the ground, 60% of this energy,
i.e., mgh × \(\frac{60}{100}\) or 0.6 mgh is converted into heat energy.
∴ Heat produced = \(\frac{0.6 m g h}{J}\)
Again, heat required to melt m g of ice = mL
(L = latent heat of fusion of ice)
∴ \(\frac{0.6 m g h}{J}\) = mL
or, h = \(\frac{J L}{0.6 g}\) (L = 80 cal ᐧ g-1 = 80 × 1000 cal ᐧ kg-1)
= \(\frac{4.2 \times(80 \times 1000)}{0.6 \times 9.8}\) = 5.71 × 104 m = 57.1 km.
Example 8.
A stirrer rotates in 1 l of water against a damping force of 0.1 N at 360 rpm in a circle of radius 5 cm. Calculate the rise in water temperature in 1 h, neglecting heat loss due to radiation. (J = 4.2 J ᐧ cal-1).
Solution:
Mass of 1 l of water, m = 1 kg.
If t is the rise in temperature of water in 1 h, then heat produced, H = mst.
Now, circumference of the circle = 2πrr; number of rotations in 1 h = 360 × 60;
∴ Displacement of the stirrer, d = 360 × 60 × 2πr;
work done, W = force × displacement
= Fd
∴ W = JH or, Fd = Jmst
or, t = \(\frac{F d}{J m s}\) = \(\frac{0.1 \times(360 \times 60 \times 2 \times \pi \times 0.05)}{4.2 \times 1 \times 1000}\)
[F = 0.1N ; r = 5 cm = 0.05 m ; s = 1000 cal ᐧ kg-1 ᐧ °C-1]
= 0.162°C.
Example 9.
10 l of water is dropped from a height of 250 m. How much heat (in calories) will be generated when the water reaches the bottom? Assuming that the entire heat will be retained by the mass of water, what will be the rise in temperature of the water? (Given J = 4.18 J ᐧ cal-1) [HS’ 08]
Solution:
Mass of 10 l of water, m = 10 kg.
Kinetic energy on impact with the ground = initial potential energy = mgh = work done (W)
So, heat generated,
H = \(\frac{W}{J}\) = \(\frac{m g h}{J}\) cal
The specific heat of water,
s = 1000 cal ᐧ kg-1 ᐧ °C-1;
If t is the rise in temperature, then
mst = H = \(\frac{m g h}{J}\)
or, t = \(\frac{g h}{J s}\) = \(\frac{9.8 \times 250}{4.18 \times 1000}\) = 0.586°C.
Example 10.
What will be the time required to heat a 15 l bucket full of water from 20°C to 40°C using a 1500W immersion heater? [HS ’06]
Solution:
Mass of 15 l of water, m = 15 kg;
specific heat of water,
s = 1000 cal ᐧ kg-1 ᐧ °C-1
Now, W = JH or, Pt = Jmsθ
where P = power of the heater = 1500W ; t = time required; θ = rise in temperature = 40 – 20 = 20°C.
So, t = \(\frac{J m s \theta}{P}\) = \(\frac{4.2 \times 15 \times 1000 \times 20}{1500}\) = 840 s = 14 min.