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Specific Heat Capacity Problem?
When resting, a person has a metabolic rate of about 4.88×105 joules per hour. The person is submerged neck-deep into a tub containing 964 kg of water at 25.48oC. If the heat from the person goes only into the water, by how much will the water temperature increase after 54 minutes of immersion?
I think I should use Q=cmdeltaT and Qloss=Qgain, but I don't know what to plug in.
Use 3500 J/(kg*C) for the specific heat of the human body and 4186 for water.
Please show steps, thanks
3 Answers
- PearlsawmeLv 710 years agoFavorite Answer
The heat given out by the person is 4.88e5 Joules per hour.
In 54 minutes the heat given out is 4.88e5*54/60 J = 439200 J
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Water absorbs this amount of heat .
Q = m c δt
439200 =964*4186* δt
δt = 0.11°C
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- Anonymous10 years ago
4.88 . 10^5 . 54/60 = 964 . 4186 . ∆T
∆T = 0.11ºC
That is all. The body is producing the heat that is going into the water. The body stays warm and the water warms up.
- Anonymous5 years ago
1. Conservation of energy. I don't know what you mean by gravitational kinetic energy however?!?! 1000 kcal = 1000 x 4180 J (there are three different definitions of calorie - but they're all fairly similar numerically) = 4.18 MJ if this is the kinetic energy of a 83 kg man 1/2 m*v^2 = 4.18 MJ v^2 = (2*4.18*10^6)/83 = 100722 v = 317 m/s 2) 3000 kcal = 12.54 MJ 3.3% of 12.54 MJ = 414 kJ mgh = 414 kJ h = 414*10^3/(60*9.81) = 702 m These are obviously silly questions and answers!!!
