A certain substance has a mass per mole of 50 g/mol. When 289 J is added as heat to a 25.5 g sample, the sample's temperature rises from 25.0 °C to 49 °C. What is the specific heat of the substance? Your answer should have units of J/kg*K, but enter only the numerical part in the box.
ANSWER::
Tf = final temperature = 490 C
TI = INITIAL TEMPERATURE = 250 C
Temperature different = 490 C - 250 C
= 240 C
M= mass per mole =50 g/mole
m= mass of substance =25.5 g = 0.0255 kg
Lets specific heat =c
then
We know that, the absorbed of heat by substance Q,
C=472.22 J/Kg*C
C=472.22 J/Kg*K [1 J/Kg.C= 1 J/Kg.K]
the specific heat of the substance IS = 472.22 J/Kg.K Only numerical value is 472.22 and unit is J/Kg.K
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