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In certain large geographic regions, the underlying rock is hot. Wells can be drilled and water circulated through the rock for heat transfer for the generation of electricity. (a) Calculate the heat transfer that can be extracted by cooling 1.00 ${km}^{3}$ of granite by $100^{\circ} {C}$. (b) How long will it take for heat transfer at the rate of 300 MW, assuming no heat transfers back into the 1.00 ${km}^{3}$ of rock by its surroundings?

$\begin{array}{ll}{\text { Part }(\mathrm{a}) :} & {Q=2.27 \times 10^{17} \mathrm{J}} \\ {\text { Part }(\mathrm{b}) :} & {t=2.10 \times 10^{5} \text { hours }}\end{array}$

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Hope College

Lectures

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in number 72. They're extracting heat from hot rocks hot granite by drilling holes and running water through, and they will know how much heat can be extracted. Um, if they run water through one kilometer, cubed and come changed a temperature, cool the granite by 100 degrees, so there's no phase change here. This is just a cuticles and civility. We know the volume, but we need the mass so we can look up the density. You look up the density of granite that was on the table, and that was 2.7 times 10 to the third kilograms per meter cube. And I know that density is mess divided by volume. So mass will be density times, will you? So I'm substituting that it for this. So I'm looking for Q. My density 2.7 times scent of 1/3. My volume, Um, this is kilometers cubed when this was meters cubed. So we used to have that. No, and it's 1000 readers, but then it's cube. So another power, Um, see, that's a specific heat. A granite. I looked it up on table 14.1, and that is 8 40 and I'm changing my temperature by 100 degrees. I multiply and I get 2.27 times 10 to the 17th. And that turns out in jewels. I'm in port de esque. If the rate no, the rate is 300 megawatts. How long would it take to get that much energy? I'm just getting my power equation power, sunken energy per time. In this case, heat per time. And in my power is 300. Mega wants migas times 10 to the sixth, the heat. I want to get us this and I want to see how much time that takes. So I took what you are about this under, and I got that. My time was, um, 7.56 times 10 to the eighth seconds. This is what's so that had to be Jules per second. Um, that's about 24 years

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