What is the specific heat capacity of tap water?

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Discussion Overview

The discussion revolves around determining the specific heat capacity of tap water and its application in calculating the time required for water to heat from 25°C to 75°C when flowing through a copper pipe placed in a 1000°C oven. Participants explore the relevant thermodynamic principles and equations necessary for this calculation.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks for the correct formula to calculate the heating time of water in a high-temperature environment.
  • Another participant suggests that Fourier's law governs the heat transfer process and notes that the heating will not be uniform throughout the water volume.
  • A participant seeks clarification on how to derive the time for the water to heat up given the heat transferred, considering the uneven heating in the pipe.
  • A detailed mathematical model is proposed, including variables such as pipe material conductivity, inner and outer radii, and the velocity of water.
  • One participant expresses the need for the specific heat capacity of tap water to complete their calculations.
  • A later reply provides a specific heat capacity value of 4200 Joules per litre per degree Celsius for tap water.

Areas of Agreement / Disagreement

Participants generally agree on the need for a mathematical approach to solve the problem, but there is no consensus on the specific methods or assumptions to be used in the calculations. The discussion remains unresolved regarding the exact implications of the uneven heating and the specific heat capacity's role in the calculations.

Contextual Notes

The discussion includes assumptions about the uniformity of heating and the dependence on various parameters such as pipe dimensions and water velocity, which are not fully resolved. The mathematical steps provided may have limitations based on these assumptions.

Ossdon
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Hi
Can anyone direct me to the correct Thermo formula to solve the following problem.
What is the time for water to heat up from 25'c to 75'c when placed in a 1000'c oven?
Tnx
Ilan
 
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Fourier's law would govern this.

The water wouldn't heat evenly, and if you want to know the time before the coldest part of the volume reaches 75 (by which time the edges will be hotter), you need to do some maths which factors in (a) the thermal conductance of the water and the container and (b) the shape of the water and thickness of the container.
 
Tnx Mikey

The furier law give me the heat transfer, but what is the equation to find the time for the water to heat up given the heat transferred?
I have water running trough a copper pipe, entering the pipe at 25°c, the pipe is placed in a 1000°c oven, I'm trying to found out what should be the length of the pipe so that the water come out of the pipe at 75°c.
I understand that the water in the pipe center will be colder then the water near the edges, but I need the average temperature of the water.

Tnx
Ilan
 
If the pipe is reasonably long in relation to diameter, I wouldn't worry too much about unequal heating.
Pipe material conductivity = k
Inner radius = r
Outer radius = R
Water temp at x from start = T(x)
Oven temp = H
Specific heat of water/unit vol = s
velocity of water = V
A = π.r^2
A full analysis gets messy because there will be heat flow along the pipe (back towards point of entry into oven) as well as radially, but let's ignore that.
Taking a slice through the pipe thickness dx, the radial heat flow depends on the temperature difference and the inner and outer radii:
F(x) = dx.2π.k.(H-T(x))/ln(R/r)
In time dt, heat entering dx of the pipe = F.dt
In that time, a volume of water V.A.dt passes any given point.
So in distance dx the water warms by dT = F/V.A.s
dT/dx = 2π.k.(H-T(x))/(ln(R/r).V.A.s)
ln((H-T)/c) = -2π.k.x/(ln(R/r).V.A.s), where c = H - T(0).
If the pipe is length L, water emerges at temp T(L) :
ln((H-T(L))/(H-T(0))) = -2π.k.L/(ln(R/r).V.A.s)

Check the math, plug in the constants and solve for L.
 
Hi again

Tnx for all the math.
I have all the constants except for the s, do you know what is the s for tap water?
didn't check yet the math.

Tnx
Ilan
 
Ossdon said:
Hi again

Tnx for all the math.
I have all the constants except for the s, do you know what is the s for tap water?
didn't check yet the math.

Tnx
Ilan

4200 Joules per litre per deg C
 

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