20 Numerical Problems — Topic-wise with Stepwise Solutions

Each problem includes clear steps and final answer. Units and symbols are shown accurately (°C, °F, ΔT, m, Q).

1. Problem 1 — Celsius to Fahrenheit:
   Convert 25 °C to Fahrenheit.
   Solution:

   Formula: T(°F) = T(°C) × 9/5 + 32

   Substitute: T(°F) = 25 × 9/5 + 32

   Calculate: 25 × 9/5 = 25 × 1.8 = 45; 45 + 32 = 77

   Answer: 25 °C = 77 °F
2. Problem 2 — Fahrenheit to Celsius:
   Convert 86 °F to Celsius.
   Solution:

   Formula: T(°C) = (T(°F) − 32) × 5/9

   Substitute: T(°C) = (86 − 32) × 5/9

   Calculate: 86 − 32 = 54; 54 × 5/9 = 54 × 0.555... = 30

   Answer: 86 °F = 30 °C
3. Problem 3 — Temperature difference:
   If the morning temperature is 18 °C and afternoon temperature is 31 °C, find the rise in temperature.
   Solution:

   ΔT = T_afternoon − T_morning = 31 − 18 = 13

   Answer: Temperature rose by 13 °C.
4. Problem 4 — Average temperature:
   If readings at 8:00 = 20 °C, 12:00 = 28 °C, 16:00 = 26 °C, find the average temperature of the day (using these samples).
   Solution:

   Average = (20 + 28 + 26) / 3 = 74 / 3 = 24.67

   Answer: ≈ 24.7 °C (rounded to one decimal place).
5. Problem 5 — Freezing point in °F:
   What is the freezing point of water (0 °C) in Fahrenheit?
   Solution:

   Use conversion: 0 × 9/5 + 32 = 32

   Answer: 0 °C = 32 °F.
6. Problem 6 — Boiling point in °F:
   What is the boiling point of water (100 °C) in Fahrenheit?
   Solution:

   100 × 9/5 + 32 = 180 + 32 = 212

   Answer: 100 °C = 212 °F.
7. Problem 7 — Mixed temperatures (simple mean):
   Two identical beakers of water have temperatures 40 °C and 20 °C. If mixed and allowed to reach equilibrium (no heat loss to surroundings), what is final temperature?
   Solution:

   Since masses and specific heat are equal and no heat loss, final T = (40 + 20) / 2 = 30 °C.

   Answer: 30 °C.
8. Problem 8 — Boiling point change with pressure (conceptual numeric):
   At high altitude a measured boiling point is 94 °C. How much lower is it than the normal boiling point (100 °C)?
   Solution:

   Difference = 100 − 94 = 6

   Answer: It is 6 °C lower than the standard boiling point.
9. Problem 9 — Convert negative temperature:
   Convert −10 °C to Fahrenheit.
   Solution:

   T(°F) = (−10) × 9/5 + 32 = −10 × 1.8 + 32 = −18 + 32 = 14

   Answer: −10 °C = 14 °F.
10. Problem 10 — Temperature reading interpolation:
    A thermometer scale marks 20 °C at one mark and 30 °C at the next. The mercury level is exactly halfway between them. What is the reading?
    Solution:

    Halfway between 20 and 30 is (20 + 30) / 2 = 25

    Answer: 25 °C.
11. Problem 11 — Effect of temperature on density (simple comparison):
    At 0 °C water is at 1.00 g/cm³. At 4 °C water reaches maximum density. If a sample of water warms from 0 °C to 10 °C, does its density increase or decrease? Explain numerically in words.
    Solution:

    Water is densest at 4 °C. Warming from 0 °C to 10 °C (past 4 °C) causes overall decrease in density compared to 4 °C. Numerically the density at 10 °C is slightly less than 1.00 g/cm³. So density decreases.

    Answer: Density decreases as temperature rises from 0 °C to 10 °C (after passing the maximum at 4 °C).
12. Problem 12 — Cooling by evaporation (conceptual numeric):
    A wet surface cools from 30 °C to 26 °C due to evaporation. What is the temperature drop ΔT?
    Solution:

    ΔT = 26 − 30 = −4 (temperature dropped by 4 °C)

    Answer: Temperature dropped by 4 °C.
13. Problem 13 — Simple percentage:
    If a region's average temperature in winter is 8 °C and expected warming raises it to 10 °C, by what percent does the temperature increase (use base 8 °C)?
    Solution:

    Increase = 10 − 8 = 2 °C. Percent increase = (2 / 8) × 100% = 25%

    Answer: 25% increase relative to 8 °C.
14. Problem 14 — Mixing unequal masses (weighted mean):
    2 kg of water at 40 °C is mixed with 1 kg of water at 20 °C. Assuming no heat loss, find final temperature (take same specific heat for both).
    Solution:

    Use weighted average: T_final = (m₁T₁ + m₂T₂) / (m₁ + m₂)

    Substitute: (2×40 + 1×20) / (2 + 1) = (80 + 20) / 3 = 100 / 3 = 33.33

    Answer: ≈ 33.3 °C.
15. Problem 15 — Using conversion in context:
    A recipe requires water at 140 °F. What is this in °C (round to 1 dp)?
    Solution:

    T(°C) = (T(°F) − 32) × 5/9 = (140 − 32) × 5/9 = 108 × 5/9 = 108 × 0.555... = 60

    Answer: 140 °F = 60 °C.
16. Problem 16 — Temperature sensor offset correction:
    A thermometer reads 2 °C higher than the true temperature (it has a +2 °C offset). If it shows 22 °C, what is the actual temperature?
    Solution:

    Actual temperature = reading − offset = 22 − 2 = 20 °C

    Answer: 20 °C.
17. Problem 17 — Boiling point and cooking time (conceptual math):
    If boiling point at sea level is 100 °C and at hill station measured as 94 °C, by what fraction does the boiling temperature decrease? Express as a percentage.
    Solution:

    Fraction decrease = (100 − 94) / 100 = 6 / 100 = 0.06 = 6%

    Answer: Boiling point decreases by 6%.
18. Problem 18 — Quick chill (salt & ice) qualitative numeric:
    When salt is added to ice, the mixture temperature drops from 0 °C to −5 °C. How much temperature change occurred?
    Solution:

    ΔT = (−5) − 0 = −5 °C (a drop of 5 °C)

    A thermometer shows marks every 2 °C. If mercury level is two small divisions above 14 °C (each small division = 0.5 °C), what is reading?

    Solution:

    Two small divisions × 0.5 °C = 1.0 °C; Reading = 14 + 1 = 15 °C.

    Answer: 15 °C.
19. Problem 20 — Simple latent heat idea (conceptual):
    A small ice cube melts at 0 °C while absorbing heat. Explain why the temperature does not rise during melting even though heat is added (no calculation needed).
    Solution:

    During melting the absorbed heat supplies the latent heat of fusion which breaks the bonds holding the solid structure together; energy is used for the change of state, not to increase temperature. So temperature remains constant at 0 °C until all ice melts.

    Answer: Temperature stays at 0 °C during melting because heat is used for the state change (latent heat), not for raising temperature.