Class 9 Physics Chapter 7 numerical problems are according to the new syllabus of the Punjab Board. Chapter 7 is related to thermal properties of matter in which you will study the working principles of how matter is converted into different substances using multiple phenomenons.
Class 9 Physics Chapter 7 Numerical Problems
The problems have been solved with easy-to-use methods using the following formulae:
°C = (°F -32) × 5/9
K = °C + 273
°F = (°C× 9/5) + 32
Numerical Problems Chapter 7 – Thermal Properties of Matter
7.1 The temperature of a normal human body on Fahrenheit scale is 98.6°F. Convert it into Celsius scale and Kelvin scale.
Solution:
Fahrenheit (°F) = 98.6°F
Celsius (°C) = ?
Kelvin (K) = ?
Use body temperature conversion:
F to C:
°C = (°F -32) × 5/9
°C = (98.6-32) × 5/9
°C= 37°C
C to K:
K = °C + 273
K = 37 + 273
K = 310 K
7.2 At what temperature Celsius and Fahrenheit thermometer reading would be the same?
Solution:
Using formula,
°C = (°F-32) × 5/9
If °C = °F
(°F-32) × 5/9 = °F
°F = -40°
7.3 Convert 5°F to Celsius and Kelvin scale.
Solution:
Convert 5°F:
Using formula,
°C = (°F-32) × 5/9
°C = (5-32) × 5/9
°C = -15°C
Using formula,
K = °C + 273
K = -15 + 273
K = 258 K
7.4 What is equivalent temperature of 25°C on Fahrenheit and Kelvin scales?
Solution:
Convert 25°C:
Using formula,
°F = (°C× 9/5) + 32
°F = (25 × 9/5) + 32
°F = 77°F
Using formula,
K = °C + 273
K = 25 + 273
K = 298 K
7.5 The ice and steam points on an ungraduated thermometer are found to be 192 mm apart. What temperature will be on Celsius scale if the length of mercury thread is at 67.2 mm above the ice point mark?
Solution:
Ungraduated thermometer:
Using proportions:
100°C : 192 mm = x°C : 67.2 mm
x°C × 192 mm = 100°C × 67.2 mm
x°C = (67.2 × 100)/192
x°C = 35°C
7.6 The length between the fixed point of liquid-in-glass thermometer is 20 cm. If the mercury level is 4.5 cm above the lower mark, what is the temperature on the Fahrenheit scale?
Solution:
Liquid thermometer:
Using proportions:
180°F : 20 cm = x°F : 4.5 cm
x°F × 20 cm = 180°F × 4.5 cm
x°F = (180 × 45)/20
x°F = 40.5°F
Temperature = 32 + 40.5 = 72.5°F
Class 9 Physics Full Book Numerical Problems with Solution
