What is the temperature change of blood?

[moonlit]

I'm having a problem solving two problems, wondering if someone can help me out...

1)Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, 0.645 kg of blood flows to the surface of the body and releases 1860 J of energy. The blood arriving at the surface has the temperature of the body interior, 37.3 °C. Assuming that blood has the same specific heat capacity as water, determine the temperature in degrees Celsius of the blood that leaves the surface and returns to the interior.

I'm assuming that you would use the equation T=Q/cm and when I do, I end up with an answer of .95 degrees celsius which I know is wrong. Can anyone explain this to me?

2)When resting, a person has a metabolic rate of about 2.34 x 105 joules per hour. The person is submerged neck-deep into a tub containing 1.81 x 103 kg of water at 26.8 °C. If the heat from the person goes only into the water, find the water temperature in degrees Celsius after half an hour.

Not really sure what equation to use here. I'm thinking Q=cmT...am I right in thinking this?

 

[turin]

Originally posted by moonlit
1)... T=Q/cm ... I end up with an answer of .95 degrees celsius ... Can anyone explain this to me?

2)... Q=cmT...am I right in thinking this?

1) Make sure that you mean ΔT in the equation. I can't figure out how to "break" the equation to get your number, though. I got something a lot closer to body temp. Hmm.

2) If you're looking for temperature, then why not use the same equation as in 1)?

Make sure you are careful with your units. They are a killer in thermo.

 

[moonlit]

For the first problem, I was wrong about it coming out to .95 degrees celsius but I still am not getting a correct answer. Here's what I did:

1860 J/[(3500 J/(kg*C)](0.645 kg) = .82 degrees celsius.

I know I'm missing a step somewhere just not sure...

 

[turin]

Originally posted by moonlit
For the first problem, ...
...
Here's what I did:

1860 J/[(3500 J/(kg*C)](0.645 kg) = .82 degrees celsius.

Whince did you get 3500 J/kg*C? Is this what you're using for the specific heat of water? My text says that the specific heat of water is 4186 J/kg*C^o at 15^oC. I am assuming (and have been told) that it doesn't change significantly from this value at 37.3^oC.

Originally posted by moonlit
I know I'm missing a step somewhere just not sure...

Yes you are. This equation will give you ΔT. You want T_f. You know T_i and ΔT. How do you find T_f from that? (BTW, you either need to intuitively realize that the temperature should decrease, or you need to be careful about the sign of Q in the equation.)