Answer:
Q = 640425 J = 640.425 KJ
Explanation:
From the law of conservation of energy:
Heat Addition = Heat Gain of Water + Heat Gain of Aluminum Kettle
[tex]Q = m_wC_w\Delta T_w + m_aC_a\Delta T_a[/tex]
where,
Q = Heat addition required = ?
[tex]m_w[/tex] = mass of water =1.8 kg
[tex]m_a[/tex] = mass of aluminum kettle = 1.1 kg
[tex]C_w[/tex] = specific heat capacity of water = = 4200 J/kg.°C
[tex]C_a[/tex] = specific heat capacity of aluminum kettle = = 890 J/kg.°C
[tex]\Delta T_w[/tex] = [tex]\Delta T_a[/tex] = change in temperature of water and kettle = 95°C - 20°C = 75°C
Therefore,
[tex]Q = (1.8\ kg)(4200\ J/kg.^oC)(75^oC)+(1.1\ kg)(890\ J/kg.^oC)(75^oC)\\[/tex]
Q = 567000 J + 73425 J
Q = 640425 J = 640.425 KJ
