Question

The temperature of a 3.0-kg material increases by 5.0°C when 6,750 J of thermal energy are added to it. What is the specific heat of the material? J/(kg-°C)​

Answer 1

450J/(kg-°C)​

Explanation:

the basic stuff of temperature is 3.0kg×5.0°C ，then assumed to be x

the thermal energy was y

then it was a functions y=kx,the k was the specific heat of the material

`k=(6750)/(15)`=450 J/kg·°C