Question

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 1

ANSWER::

T_f = final temperature = 49⁰ C

T_I = 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