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Chapter 14: Problem 23

A thermometer containing \(0.10 \mathrm{g}\) of mercury is cooled from \(15.0^{\circ} \mathrm{C}\) to \(8.5^{\circ} \mathrm{C} .\) How much energy left the mercury in this process?

Short Answer

Expert verified
Answer: Approximately 0.091 Joules of energy leaves the mercury.

Step by step solution

01

Find the change in temperature

ΔT = T_final - T_initial ΔT = 8.5°C - 15.0°C ΔT = -6.5°C The negative sign indicates that the temperature decreased, which means that the mercury lost energy.
02

Find the specific heat capacity of mercury

The specific heat capacity of mercury (\(c\)) is a known value, which is approximately \(c = 140 \frac{\mathrm{J}}{\mathrm{kg}\cdot \mathrm{K}}\). Since the mass of mercury is given in grams, we need to convert the specific heat capacity to \(\frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\) by dividing by 1000: \(c = 0.14 \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}\)
03

Calculate the heat transfer (energy left)

Now we have all the values needed to solve for the heat transfer (\(Q\)) using the formula \(Q = mcΔT\): \(Q = (0.10 \thinspace \mathrm{g}) \cdot (0.14 \thinspace \frac{\mathrm{J}}{\mathrm{g} \cdot \mathrm{K}}) \cdot (-6.5 \thinspace \mathrm{K})\) \(Q = -0.091 \thinspace \mathrm{J}\) Since the value of \(Q\) is negative, it confirms that the energy left the mercury. In this process, approximately 0.091 Joules of energy left the mercury as it cooled from 15°C to 8.5°C.

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Most popular questions from this chapter

A lizard of mass \(3.0 \mathrm{g}\) is warming itself in the bright sunlight. It casts a shadow of \(1.6 \mathrm{cm}^{2}\) on a piece of paper held perpendicularly to the Sun's rays. The intensity of sunlight at the Earth is \(1.4 \times 10^{3} \mathrm{W} / \mathrm{m}^{2},\) but only half of this energy penetrates the atmosphere and is absorbed by the lizard. (a) If the lizard has a specific heat of $4.2 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right),$ what is the rate of increase of the lizard's temperature? (b) Assuming that there is no heat loss by the lizard (to simplify), how long must the lizard lie in the Sun in order to raise its temperature by \(5.0^{\circ} \mathrm{C} ?\)
An \(83-\mathrm{kg}\) man eats a banana of energy content $1.00 \times 10^{2} \mathrm{kcal} .$ If all of the energy from the banana is converted into kinetic energy of the man, how fast is he moving, assuming he starts from rest?
It requires \(17.10 \mathrm{kJ}\) to melt \(1.00 \times 10^{2} \mathrm{g}\) of urethane $\left[\mathrm{CO}_{2}\left(\mathrm{NH}_{2}\right) \mathrm{C}_{2} \mathrm{H}_{5}\right]\( at \)48.7^{\circ} \mathrm{C} .$ What is the latent heat of fusion of urethane in \(\mathrm{kJ} / \mathrm{mol} ?\)
A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
A hotel room is in thermal equilibrium with the rooms on either side and with the hallway on a third side. The room loses heat primarily through a 1.30 -cm- thick glass window that has a height of \(76.2 \mathrm{cm}\) and a width of $156 \mathrm{cm} .\( If the temperature inside the room is \)75^{\circ} \mathrm{F}$ and the temperature outside is \(32^{\circ} \mathrm{F}\), what is the approximate rate (in \(\mathrm{kJ} / \mathrm{s}\) ) at which heat must be added to the room to maintain a constant temperature of \(75^{\circ} \mathrm{F} ?\) Ignore the stagnant air layers on either side of the glass.
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