Answer:
241 J/kg K.
Explanation:
From the complete question, the mass of the sample= 0.37 kg, Final temperature of the sample =14.5°C (287.5K), the Initial temperature of the sample = -1.2°C(271.8 K).
Therefore, the Change in temperature,∆T = final temperature of the sample - initial temperature of the sample.
Hence, ∆T= 287.5 - 271.8.
∆T= 15.7 K.
Using the formula below;
Q= MC∆T-----------------------------------(1).
Where M= mass of the sample= 0.37 kg, Q = the heat required to raise the temperature and C= specific heat capacity= 1.4 kJ= 1.4 × 1000 = 1,400 J.
Slotting in the values into equation (1) above, we have;
1400= 0.37 × C × 15.7K
1400= 5.809C.
C= 241.005336546737820.
Specific heat capacity,C;
= 241 J/kg K.
The chemist will report the specific heat capacity of the substance as 241 J/kg K.
