Answer:Explained
Explanation:
Given
mass of aluminium kettle([tex]m_k[/tex])=1.05 kg
(a)Heat capacity of the kettle=Heat capacity of Aluminium
[tex]C_{Al}=0.9 J/gm-K[/tex]
(b)Heat is required to increase the temperature of this kettle from [tex]23^{\circ} to 99 ^{\circ}[/tex]
[tex]Q=m_kc_k\Delta T[/tex]
[tex]Q=1.05\times 0.9\times 1000\times \left ( 99-23\right )[/tex]
Q=71.82 kJ
(c)If kettle contains water 1.25 L of water i.e. 1.246 kg of water then heat required to raise the temperature from 23 to 99
[tex]Q=m_kc_k\left ( \Delta T\right )+m_wc_w\left ( \Delta T\right )[/tex]
[tex]Q=1.05\times 0.9\times 1000\times \left ( 99-23\right )+1.246\times 4.184\times 1000\times \left ( 99-23\right )[/tex]
Q=71.82+396.20=468.028 kJ
