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Chapter 15: Problem 1

On a cold day, Ming rubs her hands together to warm them up. She presses her hands together with a force of \(5.0 \mathrm{N} .\) Each time she rubs them back and forth they move a distance of \(16 \mathrm{cm}\) with a coefficient of kinetic friction of \(0.45 .\) Assuming no heat flow to the surroundings, after she has rubbed her hands back and forth eight times, by how much has the internal energy of her hands increased?

Short Answer

Expert verified
total work done by the friction force. This is because the work done by friction turns into heat energy, which increases the internal energy of her hands. Thus: ΔE_internal = W_total = 2.88 J Therefore, the increase in internal energy of Ming's hands is 2.88 J.

Step by step solution

01

Calculate the friction force

First, we should calculate the friction force between Ming's hands. The friction force can be calculated using the equation \(F_{friction} = μ * F_n\), where \(F_{friction}\) is the friction force, \(μ\) is the coefficient of kinetic friction, and \(F_n\) is the normal force. In this case, the normal force is the force Ming applies on her hands, which is 5.0 N, and the coefficient of kinetic friction is 0.45. Thus: \(F_{friction} = μ * F_n = 0.45 * 5.0 \mathrm{N} = 2.25 \mathrm{N}\)
02

Calculate the work done by friction force

Now that we have the friction force, we can calculate the work done by this force each time Ming rubs her hands back and forth. The work done by a force acting over a distance can be calculated using the equation \(W = F * d\), where \(W\) is the work done, \(F\) is the force, and \(d\) is the distance. In this case, the force is the friction force, which is 2.25 N, and the distance is 16 cm (0.16 m). Thus: \(W = F * d = 2.25\mathrm{N} * 0.16\mathrm{m} = 0.36\mathrm{J}\)
03

Calculate the total work done

To find the total work done after Ming rubs her hands back and forth eight times, we just need to multiply the work done each time (0.36 J) by the number of times she does it (8). \(W_{total} = W * n = 0.36\mathrm{J} * 8 = 2.88\mathrm{J}\)
04

Determine the increase in internal energy

Assuming no heat flow to the surroundings, the increase in internal energy of Ming's hands is equal to the

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Most popular questions from this chapter

List these in order of increasing entropy: (a) \(0.5 \mathrm{kg}\) of ice and \(0.5 \mathrm{kg}\) of (liquid) water at \(0^{\circ} \mathrm{C} ;\) (b) $1 \mathrm{kg}\( of ice at \)0^{\circ} \mathrm{C} ;\( (c) \)1 \mathrm{kg}$ of (liquid) water at \(0^{\circ} \mathrm{C} ;\) (d) \(1 \mathrm{kg}\) of water at \(20^{\circ} \mathrm{C}\).
Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$
A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
Show that the coefficient of performance for a reversible heat pump is $1 /\left(1-T_{\mathrm{C}} / T_{\mathrm{H}}\right)$.
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