Question

What is the specific heat of a substance that absorbs 228 joules of heat when a sample of 706 g of the substance increases in tempature from 26.0 c to 88.8 c

Answer 1

Answer:

$c=5.14\ J/kg ^{\circ} C$

Explanation:

Given that,

Heat absorbed by a sample, Q = 228 J

Mass of a sample, m = 706 g = 0.706 kg

Initial temperature is 26 °C and final temperature is 88.8°C

We need to find the heat absorbed by the sample. The heat absorbed by an object is given by :

$Q=mc\Delta T\\\\\text{Where c is specific heat of sample}\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{228\ J}{0.706\ kg(88.8-26)^{\circ} C}\\\\c=5.14\ J/kg ^{\circ} C$

So, the specific heat of the sample is $5.14\ J/kg ^{\circ} C$.