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

Answer:

$c=10\ J/kg^{\circ} C$

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 :

$Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C$

So, the specific heat of the object is $10\ J/kg^{\circ} C$.