Water's specific heat in the range from 0^∘ C to 100^∘ C varies with temperature according to th

Water's specific heat in the range from 0^∘ C to 100^∘ C...

Water's specific heat in the range from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ varies with temperature according to the equation $c(T)=c_{0}+a T+b T^{2},$ where $c_{0}=4207.9 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, a=-1.292 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}^{2},$ and $b=0.01330 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}^{3}$ Use this expression to find the heat required to raise the temperature of $1.000 \mathrm{~kg}$ of water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. By what percentage does this differ from the result you would get using the value of $c$ in Table 16.1 over the entire temperature range?

Water's specific heat in the range from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ varies with temperature according to the equation $c(T)=c_{0}+a T+b T^{2},$ where $c_{0}=4207.9 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, a=-1.292 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}^{2},$ and $b=0.01330 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}^{3}$
Use this expression to find the heat required to raise the temperature of $1.000 \mathrm{~kg}$ of water from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. By what percentage does this differ from the result you would get using the value of $c$ in Table 16.1 over the entire temperature range?

Key Concepts

Specific Heat Capacity

This concept refers to the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It is a fundamental property in thermodynamics and plays a crucial role in understanding how materials absorb and transfer heat energy.

Temperature-Dependent Specific Heat

Often, the specific heat of a substance is not constant and can vary with temperature. This variation must be taken into account for precise calculations of energy transfer over a range of temperatures, especially for substances like water whose specific heat can be expressed as a function of temperature.

Integration for Variable Specific Heat

When the specific heat capacity varies with temperature, the total heat required to change the temperature of a substance is determined by integrating the temperature-dependent specific heat function over the given temperature range. This mathematical approach sums the infinitesimal amounts of heat added over each differential temperature interval.

Percentage Difference Calculation

After calculating the heat required using both the variable specific heat function and a constant average value, the percentage difference provides a quantifiable measure of how much the two results diverge. This is important in evaluating the impact of assuming constant properties versus accounting for temperature variability.

Transcript

00:01 This problem, we will be calculating the amount of energy required to raise the temperature of water from 0 degrees centigrade to 100 degrees centigrade.

00:10 However, the specific heat c of water this time is not a constant, instead it changes with temperature.

00:18 So we'll need to differentiate and integrate and use some calculus.

00:22 So the formula that we will be using is delta q is equal to m, c, delta t, where m, c, delta, t.

00:32 M is the mass, c is the specific heat, delta t is change in temperature, and delta q is the required heat.

00:48 So what we can do is think about in terms of differentials.

00:52 This is a change, and this is a change.

00:55 This is a change in heat, and this is a change in temperature.

01:00 So if you take a small change and call it dq, then it will result in a small change.

01:12 The small change dq will be caused by a small change in temperature dt.

01:19 Now we can go ahead and substitute in the values.

01:23 So the mass is 1.

01:25 And c is this entire expression over here.

01:29 So 4207 .9 minus 1 .292 .t plus 0 .032t plus 0 .133.

01:46 0 t squared the whole multiply by dd now to get q we'll need to integrate this entire expression over our required range this time there's a temperature and our temperature goes from 0 to 100 that means our required heat energy is simply the integral from 0 to 100 of 4207 .9 minus 1 .2 .29 minus 1 .229 plus 0 .01 .3 .0 t squared d t.

02:30 Now it's a relatively simple integral.

02:33 So the integral of 42 of 7 .9 is simply 42 of 7 .9 t.

02:40 The second term, the integral is just...

02:47 So we're using the power rule.

02:48 So whenever we have a polynomial, so the integral of x to the n is always equal to the integral of x to the n plus 1...

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