Question

c) An aluminum tea kettle with mass 1.10 kg containing 1.80 kg of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20o C to 95o C

Answer 1

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

$Q = m_wC_w\Delta T_w + m_aC_a\Delta T_a$

where,

Q = Heat addition required = ?

$m_w$ = mass of water =1.8 kg

$m_a$ = mass of aluminum kettle = 1.1 kg

$C_w$ = specific heat capacity of water = = 4200 J/kg.°C

$C_a$  = specific heat capacity of aluminum kettle = = 890 J/kg.°C

$\Delta T_w$ = $\Delta T_a$ = change in temperature of water and kettle = 95°C - 20°C = 75°C

Therefore,

$Q = (1.8\ kg)(4200\ J/kg.^oC)(75^oC)+(1.1\ kg)(890\ J/kg.^oC)(75^oC)$

Q = 567000 J + 73425 J

Q = 640425 J = 640.425 KJ