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1. Given the specific heat of lead is 0.129 J/g.C and that it takes 93.4J of energy to
heat a sample of lead from 22.3°C to 40.4°C find the mass of the lead.
m=

Respuesta :

olatunbosunadeoke

Answer: 40 grams

Explanation:

The quantity of Heat Energy (Q) required to heat a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since Q = 93.4J

M = ?

C = 0.129 J/g.C

Φ = 40.4°C - 22.3°C = 18.1°C

Then, Q = MCΦ

Make Mass, M the subject formula

M = Q/CΦ

M = (93.4J) / (0.129 J/g.C x 18.1°C)

M = 93.4J / 2.33J/g

M = 40 g

Thus, the mass of the lead is 40 grams

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