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Question
1300 J of heat energy is supplied to raise the temperature of 6.5 kg of lead from 20° C to 40°C. Calculate the specific heat capacity of lead.
Solution
Heat energy supplied (Q) = 1300 J
Mass of lead (m) = 6.5 kg
Change in temperature (Δ t) = (40 - 20)°C = 20° C (or 20 K)
Specific heat capacity (c) = ?
From relation, Q = `m xx c xx Delta t`
Substituting the values in the formula above we get,
1300 = 6.5 × c × 20
`=> c = 1300/(6.5 xx 20)`
`=> c = 1300/130`
⇒ c = 10 J kg-1 k-1
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