500 gram sample of an unknown metal releases 640 Joules as it cools from 55.0 oC to 250 oC. What is the specific heat of the sample?

c = 0.043 j/ g. °C

Explanation:

Given data:

Amount of sample = 500 g

Initial temperature = 55.0°C

Final temperature = 25.0°C

Heat required released = 640 J

Specific heat capacity of sample = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 25°C - 55°C

ΔT = -30°C

by putting values,

-640 J = 500 g × c × -30 °C

c  = 640 J / 500 g× 30 °C

c = 640 J /15000 g. °C

c = 0.043 j/ g. °C

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