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What is the mass of a sample whose specific heat is 6.28 J/g·°C if the sample absorbs 19.4 J and has a temperature increase of 22.9 °C?

a)0.135 g
b)0.324 g
c)7.41 g
d)70.7 g

Respuesta :

We have: Q = m.s.Δt
m = Q / s.Δt
Here, Q = 19.4 J
s = 6.28 J/g C
Δt = 22.9

Substitute their values into the expression:
m = 19.4 / 6.28×22.9
m = 19.4 / 143.81
m = 0.135 g

In short, Your Answer would be Option A

Hope this helps!

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