Question

37.9 grams of an unknown substance undergoes a temperature increase of

25.0*C after absorbing 969 J. What is the specific heat of the substance?

Answer 1

1.023 J / g °C

Step-by-step explanation:

m = 37.9 grams

ΔT = 25.0*C

H = 969 J

c = ?

The equation relating these equation is;

H = mcΔT

making c subject of formulae;

c = H / mΔT

c = 969 J / (37.9 g * 25.0*C)

Upon solving;

c = 1.023 J / g °C