Calculate: A. What is the specific heat of plastic when 440 Joules of heat are radiated on a 140 gram slab of plastic which changes temperature from 20 -22 oC?

1.571 J/g°C

Explanation:

Using Q = mc∆T

Where;

Q = amount of heat in Joules

m = mass of substance (grams)

c = specific heat capacity

∆T = change in temperature (°C)

According to the question;

m = 140g, Q = 440J, ∆T = 22°C - 20°C = 2°C

Using Q = mc∆T

c = Q/m∆T

c = 440/ 140 × 2

c = 440/280

c = 1.571 J/g°C

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