Physics
1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from 20° C to 40° C. Calculate the specific heat capacity of lead.
Calorimetry
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Answer
Given,
Heat energy supplied (Q) = 1300 J
Mass of lead (m) = 0.5 kg
Change in temperature (△t) = (40 – 20)°C = 20° C
Specific heat capacity (c) = ?
From relation,
Substituting the values in the formula above we get,
Hence, specific heat capacity of lead = 130 J kg-1 K-1
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