Question

What is the specific heat of an object if it takes 1200 J to raise the temperature of a 20 kg object by 6 degrees C?

Answer 1

`c=10\ J/kg^(\circ) C`

Step-by-step explanation:

Given that,

Heat required, Q = 1200 J

Mass of the object, m = 20 kg

The increase in temperature,
`\Delta T=6^(\circ) C`

We need to find the specific heat of the object. The heat required to raise the temperature is given by :

`Q=mc\Delta T\\\\c=(Q)/(m\Delta T)\\\\c=(1200)/(20* 6)\\\\c=10\ J/kg^(\circ) C`

So, the specific heat of the object is
`10\ J/kg^(\circ) C`.