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jonny [76]
3 years ago
9

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 surround

ings, how much heat must be added to raise the temperature from 20o C to 95o C
Physics
1 answer:
True [87]3 years ago
8 0

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

<u>Q = 640425 J = 640.425 KJ</u>

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The second stone hits the ground exactly one second after the first.

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t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

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