The specific heat of lead is 0.030 cal/gC° 300g of lead shot at 100°C is mixed with 100 g of water at 70°C in an insulated container. The final temperature of the mixture is?

Answer: 72.5°C

To find the final temperature, we can use the principle of heat transfer:

Heat lost by lead = Heat gained by water

Given:

Specific heat of lead (c Pb) = 0.030 cal/g°C

Mass of lead (m Pb) = 300 g

Initial temperature of lead (T Pb) = 100°C

Mass of water (m w) = 100 g

Initial temperature of water (T w) = 70°C

Specific heat of water (c w) = 1 cal/g°C (approximately)

Let the final temperature be T f.

Heat lost by lead = m Pb × c Pb × (T Pb - T f)

= 300 × 0.030 × (100 - T f)

Heat gained by water = m w × c w × (T f - T w)

= 100 × 1 × (T f - 70)

Equating the two expressions:

300 × 0.030 × (100 - T f) = 100 × 1 × (T f - 70)

Simplifying the equation:

9(100 - T f) = 100T f - 7000

900 - 9T f = 100T f - 7000

109T f = 7900

T f ≈ 72.5°C

This question appeared in Past Papers (5 times)

Lecturer Physics Past Papers and Syllabus (1 times)

SPSC 25 Years Past Papers Subject Wise (Solved) (2 times)

SPSC Past Papers (2 times)

This question appeared in Subjects (1 times)

EVERYDAY SCIENCE (1 times)

Related MCQs