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?

Explanation

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

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