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# Heat content or Btu value of water at various temperatures?

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Help??

Where can I find the Btu value or heat content of water at various temperatures such as 50, 60, 70, 80, 90 & 100 degrees F? Say 1 lb of water.

Is it also generally safe to say that if the water is in contact with another surface, that the water will give up its heat content so long as the adjoining mass is cooler than the water??

For instance, if a 1 lb block of wood at 65 degrees F is submerged in water that is 80 degrees F, how much heat will flow from the water to the wood?

How does the thermal conductivity of the wood play into this?? Or does thermal conductivity relate only to the passage of gases through a substrate?

https://brainmass.com/physics/internal-energy/heat-content-btu-value-water-various-temperatures-151333

## SOLUTION This solution is FREE courtesy of BrainMass!

(a)
I believe what you actually need is not the total energy in a pound of water but the change of this energy with the change of temperature.

This is addressed by heat capacity which in turn is determined by specific heat.
The specific heat of water varies very little between the temperatures you mention, so you can take it to be the same and its value is

c = 1 Btu / (Lb*ºF)

(see e.g. http://www.refrigerationbasics.com/1024x768/definitions1.htm).

The fact that it is exactly 1 is not a coincidence, because the heat capacity of 1 pound of water per ºF serves as a DEFINITION of Btu.

Given some mass m of water, to change its temperature by some &#916;T, we need the amount of energy

&#916;Q = c*m* &#916;T.

For instance, to raise the temperature of 1 pound of water from 50ºF to 60ºF, we would need to input

( 1 Btu/LbºF ) * 1 Lb * 10ºF = 10 Btu

of heat energy into the water.

Conversely, if 1 Lb of water cools from 60ºF to 50ºF it gives off 10 Btu, or if it cools from 100ºF to 50ºF it gives off 50 Btu.

(b)
It is true that if two bodies of different temperatures are put in thermal contact with each other and isolated from the rest of the world, they will exchange heat energy until they both have the same temperature.

In particular, if warmer water is in contact with cooler wood, the water will give heat to wood until their temperatures become the same.

(c)
Suppose a body of mass m_1, specific heat c_1, and temperature T_1 is in thermal contact with another body of mass m_2, specific heat c_2, and temperature T_2, and the two are left alone isolated from the rest of the world.

They will come to thermal equilibrium at temperature T which is determined from conservation of energy:
The heat loss of one of them is the heat gain of the other, or it also can be said that the sum of changes in their internal energies must be zero:

0 = c_1*m_1*(T-T_1) + c_2*m_2*(T-T_2) (1)

From equation (1) we find that

T = ( c_1*m_1*T_1 + c_2*m_2*T_2 ) / ( c_1*m_1 + c_2*m_2 ) (2)

and the amount of heat that was passed from one body to another is

&#916;Q = c_1*m_1*|T-T_1| = c_2*m_2*|T-T_2| (3)

(we write here absolute values as we do not know yet which is larger, T_1 or T_2).

In your question (c) you know the initial temperatures of water and wood,
water specific heat is 1 Btu / (Lb*ºF) and wood specific heat can be found either in your sources or on the web.
For instance the web page cited above says that wood specific heat can be between 0.32 and 0.48 Btu / (Lb*ºF).
You also have the mass of the wood there and all that is missing is the mass of the water.
When you know all the input values, you can use equation (3) to find the amount of heat transferred and equation (2) to find the equilibrium temperature.

(d)
The thermal conductivity of the wood as well as that of the water determine the rate at which their temperatures equilibrate.
The passage of gases through a substrate is determined by diffusion coefficient.
The gas diffusion can contribute to thermal conductivity but it would not be all of it and in any case these are two quite different things not to be confused.

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