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A bucket has 101 water at 15°c. How much water at 50°c should be added to get a mixture of temperature 40°C? (Assume no heat loss in mixing.)
Question

A bucket has 101 water at 15°c. How much water at 50°c should be added to get a mixture of temperature 40°C? (Assume no heat loss in mixing.)

A.

15 L

B.

20 L

C.

25 L

D.

30 L

Correct option is C

Given:
Initial water: 10 L at 15°C
Added water: x L at 50°C
Final mixture temperature: 40°C
No heat loss (heat gained = heat lost).
Concept Used:
The principle of heat exchange states:

Heat gained by cooler water = Heat lost by hotter water

The heat gained or lost by a substance is given by:

Q = m × c × ΔT

where:

m = mass (or volume in this case, since density of water is 1 kg/L)
c = specific heat (same for water, so it cancels out)
ΔT = change in temperature
Thus, we set up the equation:

(Mass of cold water) × (Temp increase) = (Mass of hot water) × (Temp decrease)

Solution:

Heat gained by cold water (10 L at 15°C):
10 × (40 - 15) = 10 × 25 = 250

Heat lost by hot water (x L at 50°C):
x × (50 - 40) = x × 10 = 10x

Setting heat gained = heat lost:

250 = 10x

Solving for x:

x = 250 / 10 = 25 L

Final Answer:
(C) 25 L

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