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

`c=5.14\ J/kg ^(\circ) C`

Step-by-step 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=(Q)/(m\Delta T)\\\\c=(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`.