The temperature on Celsius scale is 25ºC. What is the corresponding temperature on the Fahrenheit scale ?
To convert from Celsius to Fahrenheit, we use the formula:
\[
F = \frac{9}{5}C + 32
\]
Given \( C = 25^\circ \):
\[
F = \frac{9}{5} \times 25 + 32
\]
\[
F = 45 + 32 = 77
\]
So, the corresponding temperature is 77ºF.
If a thermometer reads freezing point of water as 20ºC and boiling point as 150ºC, how much thermometer read when the actual temperature is 60ºC ?
To solve this, we can set up a linear relationship between the actual Celsius scale (0ºC to 100ºC) and the thermometer's faulty scale (20ºC to 150ºC).
1. Set up the linear equation:
The faulty thermometer's scale can be represented as:
\[
T_{\text{faulty}} = a \cdot T_{\text{actual}} + b
\]
Using the freezing point:
\[
20 = a \cdot 0 + b \Rightarrow b = 20
\]
Using the boiling point:
\[
150 = a \cdot 100 + 20
\]
\[
130 = 100a \Rightarrow a = 1.3
\]
So, the relation is:
\[
T_{\text{faulty}} = 1.3 \cdot T_{\text{actual}} + 20
\]
2. Find the faulty reading at 60ºC actual temperature:
\[
T_{\text{faulty}} = 1.3 \cdot 60 + 20 = 78 + 20 = 98
\]
Therefore, the thermometer will read 98ºC at an actual temperature of 60ºC.
Two temperature scales A and B are related by: \[ \frac{A-42}{100}=\frac{B-7}{220}\]
At which temperature two scales have the same reading ?
Given the relation between temperature scales \( A \) and \( B \):
\[
\frac{A - 42}{100} = \frac{B - 7}{220}
\]
To find the temperature at which both scales show the same reading, set \( A = B = x \):
\[
\frac{x - 42}{100} = \frac{x - 7}{220}
\]
Cross-multiplying:
\[
220(x - 42) = 100(x - 7)
\]
Expanding and solving for \( x \):
\[
220x - 9240 = 100x - 700
\]
\[
120x = 8540
\]
\[
x = \frac{8540}{120} = 12
\]
Thus, the temperature at which both scales read the same is 12.
A rod of length 20 cm is made of metal. It expands by 0.075 cm when its temperature is raised from 0ºC to 100ºC. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature. A third rod of the same length is composed of two parts, one of metal A and the other of metal B. This rod expands by 0.060 cm for the same change in temperature. The portion made of metal A has the length :
Let the length of the part made of metal A be \( L_A \) and that of metal B be \( L_B \), with \( L_A + L_B = 20 \) cm.
Given:
- Expansion of metal A's rod = 0.075 cm, so expansion per cm for metal A = \( \frac{0.075}{20} = 0.00375 \) cm.
- Expansion of metal B's rod = 0.045 cm, so expansion per cm for metal B = \( \frac{0.045}{20} = 0.00225 \) cm.
The combined expansion of the third rod is 0.060 cm:
\[
L_A \cdot 0.00375 + L_B \cdot 0.00225 = 0.060
\]
Since \( L_B = 20 - L_A \), substitute:
\[
L_A \cdot 0.00375 + (20 - L_A) \cdot 0.00225 = 0.060
\]
Expanding and solving:
\[
0.00375L_A + 0.045 - 0.00225L_A = 0.060
\]
\[
0.0015L_A = 0.015
\]
\[
L_A = \frac{0.015}{0.0015} = 10 \text{ cm}
\]
So, the portion made of metal A has length 10 cm.
A cylindrical metal rod of length L0 is shaped into a ring with a small gap as shown. On heating the system.
On heating an object its photographic expansion takes place. so, distance between any two point increases.
The absolute zero temperature in Fahrenheit scale is :
To find absolute zero in Fahrenheit, we use the relationship between Celsius and Fahrenheit:
\[
F = \frac{9}{5}C + 32
\]
Absolute zero in Celsius is \(-273.15^\circ C\). Substitute this into the formula:
\[
F = \frac{9}{5}(-273.15) + 32
\]
\[
F = -491.67 + 32 = -459.67
\]
Rounding to the nearest whole number, we get \(-460^\circ F\).
In some old science notes we come across a temperature scale called Z scale (gargi scale) on which boiling point of water is 65ºZ and freezing point is –14ºZ. It is found that a change of 1º on Z scale corresponds to change of xº on Fahrenheit scale. Then x is :
To find \( x \), we need to compare the Z scale with the Fahrenheit scale.
1. Range on the Z scale:
\[
65^\circ Z - (-14^\circ Z) = 65 + 14 = 79^\circ Z
\]
2. Range on the Fahrenheit scale:
The boiling and freezing points of water are 212ºF and 32ºF, respectively, so:
\[
212^\circ F - 32^\circ F = 180^\circ F
\]
3. Relationship between scales:
A change of \( 79^\circ Z \) corresponds to \( 180^\circ F \). Therefore:
\[
x = \frac{180}{79}
\]
So, \( x = \frac{180}{79} \).
A bar of iron is 10 cm at 20ºC. At 19ºC it will be (α of iron = 11 × 10–6/ºC)
The change in length \( \Delta L \) due to temperature change is given by:
\[
\Delta L = L_0 \alpha \Delta T
\]
where:
- \( L_0 = 10 \, \text{cm} \)
- \( \alpha = 11 \times 10^{-6} / ^\circ \text{C} \)
- \( \Delta T = 20^\circ \text{C} - 19^\circ \text{C} = 1^\circ \text{C} \)
Substitute the values:
\[
\Delta L = 10 \times 11 \times 10^{-6} \times 1 = 11 \times 10^{-5} \, \text{cm}
\]
Thus, the bar will be \( 11 \times 10^{-5} \, \text{cm} \) shorter at 19ºC.