1) Aim
2) Apparatus
3) Diagram
4) Formula with explanation/law
5) Theory
6) Observations
7) Observation Table
8) Calculation
9) Result/Conclusion
10) Precautions
11) Questions
12) Graph
आता आपण Experiment चे Solutions पाहण्यास सुरुवात करुया :-
Class 12 Physics Practical No 3 : Newton's law of cooling Solutions Maharashtra Board
[Page No 23]
Aim :
To study the relationship between the temperature of a hot body and time by plotting a cooling curve.
Apparatus :
A thimble or calorimeter, Constant temperature enclosure, a thermometer, stop clock, hot water bath etc.
Newton's law of cooling :
The rate of fall of temperature (dθ/dt) by a body is directly proportional to the difference of temperature (θ-θθ) of the body over the surroundings, provided the difference is small.
Diagram :
Theory :
The rate at which a hot body loses heat is directly proportional to the difference between the temperature of the hot body and that of its surroundings and depends on the nature of material and the surface area of the body. This is newton's law of cooling.
Rate of loss of heat dθ/dt = k(θ-θθ)
where, K = is the constant of proportionality K = k'/ms
also, dθ/dt = ms dθ/dt
∴ ms dθ/dt = -k'(θ-θθ)
∴ dθ/dt = -k'/ms * (θ-θθ)
Formula :
dθ/dt = -k(θ-θθ)
K' = k/ms is another constant it is water equivalent of the calorimeter negative fig indicates that loss of heat implies temperature decrease.
[Page No 24]
Observations :
1) Least count of stop watch = 0.1 s.
2) Least count of thermometer = 1 ℃.
3) Temperature of the surrounding (θθ) = 25 ℃.
Observation Table (I) :
| Obs. No. |
Time in min | Temperature of water ℃ |
Obs. No. |
Time in min | Temperature of water ℃ |
|---|---|---|---|---|---|
| 1 | 0 | 72 | 16 | 15 | 57 |
| 2 | 1 | 71 | 17 | 16 | 57 |
| 3 | 2 | 69 | 18 | 17 | 56 |
| 4 | 3 | 68 | 19 | 18 | 55 |
| 5 | 4 | 67 | 20 | 19 | 55 |
| 6 | 5 | 66 | 21 | 20 | 54 |
| 7 | 6 | 65 | 22 | 21 | 54 |
| 8 | 7 | 64 | 23 | 22 | 53 |
| 9 | 8 | 63 | 24 | 23 | 52 |
| 10 | 9 | 62 | 25 | 24 | 52 |
| 11 | 10 | 61 | 26 | 25 | 52 |
| 12 | 11 | 61 | 27 | 26 | 51 |
| 13 | 12 | 58 | 28 | 27 | 50 |
| 14 | 13 | 58 | 29 | 28 | 50 |
| 15 | 14 | 58 | 30 | 29 | 49 |
[Page No 25]
Observation Table From cooling (II) : θθ = 25 ℃ (Room temp.)
| Obs. No. |
Temp θ℃ | dθ/dt ℃/min | (θ-θθ) |
|---|---|---|---|
| 1 | 69.35 | 1.32 | 44.35 |
| 2 | 64.65 | 1.025 | 39.65 |
| 3 | 60.9 | 0.085 | 35.9 |
| 4 | 57.6 | 0.8 | 32.6 |
| 5 | 54.9 | 0.55 | 29.9 |
[Page No 26]
Result :
1) The nature of the graph of temperature against time is a curve. It is known as Newton's cooling curve.
2) This cooling curve will be steep at first, after it will become less steep as the temperature approaches to temperature of the surrounding.
3) As the graph of dθ/dt against (θ-θθ) is straight line, Passing through origin Newton's law of cooling is verified.
4) Room temperature by graph, θθ = 26 ℃
Precautions :
i) Make sure that the bulb of thermometer is well inside the water.
ii) The enclosure should have proper insulation to avoid loss of heat due to conclusion or convection from hot water.
iii) The water in the calorimeter should be gently stirred continually.
Questions :
1) Certain quantity of water cools from 70℃ to 60℃ in the first 5 mint and 54℃ in the next 5 min. Then what well be the temperature of the surrounding (Ans. 45℃)
Ans : -k(65-T)
-k(57-T)
60-70 = -10
-10/-6 = (65-T)/(57-T)
∴ 285-5T = 195-3T
∴ 90 = 2T
∴ T = 45℃
2) You take an ice-cream out of the freezer, kept at -18℃. outside it is 32℃ After one minute, the ice cream has warmed to -8℃ what is the temperature of the ice-cream after five minutes? (Ans. ≈ 15.6℃).
Ans : for t=1 min.
-8 = 32 + (-18-32)*e-k
∴ e-k = (32+8)/50
e-k = 40/50
e-k = 0.8
ln(e-k ) = ln(0.8)
-k = ln(0.8)
multiply by -1 to find k
k = -ln(0.8) ≈ 0.2231
T(5) = 32+(-18-32)*e(0.2231*5)
T(5) = 32-50*0.3274
T(5) = 32-16.37
T(5) = 15.63℃
3) State Newton's law of cooling. write its expression in the form of specific heat.
Ans : The rate of fall of temperature (dθ/dt) by body is directly proportional to the difference of temperature (θ-θθ) of the body over the surroundings, provided the difference is small.
dθ/dt = -k(θ-θθ).
Graphs :
1) Cooling curve
2) Graph of (dθ/dt) versus θ
3) Graph of (dθ/dt) versus (θ-θθ)
अशा प्रकारे मित्रांनो आपण इयत्ता बारावी Physics Practical Number 3 - Newton's law of cooling याचे Solutions पहिले.
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