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Chapter 14: Problem 84

A spring of force constant \(k=8.4 \times 10^{3} \mathrm{N} / \mathrm{m}\) is compressed by \(0.10 \mathrm{m} .\) It is placed into a vessel containing $1.0 \mathrm{kg}$ of water and then released. Assuming all the energy from the spring goes into heating the water, find the change in temperature of the water.

Short Answer

Expert verified
Answer: The change in temperature of the water is approximately 0.01005 K.

Step by step solution

01

Calculate the potential energy of the spring

First, we need to calculate the potential energy stored in the compressed spring using the formula \(PE_{spring} = \frac{1}{2}kx^2\), where \(k\) is the force constant and \(x\) is the compression distance. We have: \(k = 8.4 \times 10^{3} \mathrm{N/m}\) and \(x = 0.10 \mathrm{m}\). So, \(PE_{spring} = \frac{1}{2} \times (8.4 \times 10^{3} \mathrm{N/m}) \times (0.10 \mathrm{m})^2\)
02

Compute the potential energy of the spring

Now, let's compute the potential energy of the spring. \(PE_{spring} = \frac{1}{2} \times (8.4 \times 10^{3} \mathrm{N/m}) \times (0.10 \mathrm{m})^2 = 42 \mathrm{J}\)
03

Apply energy conservation and find the heat energy gained by water

Now, we apply the conservation of energy principle and equate the potential energy of the spring to the heat energy gained by the water. So, \(PE_{spring} = Q\) where \(Q = mc\Delta T\). We are given: \(m = 1.0 \mathrm{kg}\), and the specific heat capacity of water is \(c = 4.18 \times 10^3 \mathrm{J/(kg\cdot K)}\). Therefore, \(42 \mathrm{J} = (1.0 \mathrm{kg}) \times (4.18 \times 10^3 \mathrm{J/(kg\cdot K)})\Delta T\)
04

Calculate the change in temperature of the water

Now, we can solve for the change in temperature, \(\Delta T\). \(\Delta T = \frac{42 \mathrm{J}}{(1.0 \mathrm{kg}) \times (4.18 \times 10^3 \mathrm{J/(kg\cdot K)})} = 0.01005 \mathrm{K}\) So, the change in temperature of the water is approximately \(0.01005 \mathrm{K}\).

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Most popular questions from this chapter

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