Ok, I have worked this one multiple times and can\'t get the right answer: An al

Question

Ok, I have worked this one multiple times and can't get the right answer:

An aluminum tea kettle with mass 1.45 kg and containing 1.90 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 19.0 C to 80.0 C ?

Here is what I have done:

Kettle = mass times coefficient = 1.45*.897 = 1300.65
Water = 1.90*4.186 = 7954.4
Add them = 9254.61
Then multiply times the change in temp which is 61 = 564.5 but that is not correct. Any help would be appreicated.

Explanation / Answer

we know, specific heat of aluminum, C_alunimum = 0.897 J/(g C) = 897 J/(kg C)

specific heat of water, C_water = 4.186 J/(g C) = 4186 J/(kg C)

change in temperature, T2 - T1 = 80 - 19

= 61 C

m_aluminum = 1.45 kg

m_water = 1.9 kg

heat must be added = m_alumimum*C_alumnium*(T2-T1) + m_water*C_water*(T2-T1)

= 1.45*897*61 + 1.9*4186*61

= 564497 J

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