Answer:
[tex]c=10\ J/kg^{\circ} C[/tex]
Explanation:
Given that,
Heat required, Q = 1200 J
Mass of the object, m = 20 kg
The increase in temperature, [tex]\Delta T=6^{\circ} C[/tex]
We need to find the specific heat of the object. The heat required to raise the temperature is given by :
[tex]Q=mc\Delta T\\\\c=\dfrac{Q}{m\Delta T}\\\\c=\dfrac{1200}{20\times 6}\\\\c=10\ J/kg^{\circ} C[/tex]
So, the specific heat of the object is [tex]10\ J/kg^{\circ} C[/tex].
