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Chapter 14: Problem 95

A 2.0 -kg block of copper at \(100.0^{\circ} \mathrm{C}\) is placed into $1.0 \mathrm{kg}$ of water in a 2.0 -kg iron pot. The water and the iron pot are at \(25.0^{\circ} \mathrm{C}\) just before the copper block is placed into the pot. What is the final temperature of the water, assuming negligible heat flow to the environment?

Short Answer

Expert verified
Answer: The final temperature of the water is approximately \(7.68^{\circ}C\).

Step by step solution

01

List the known values

- Mass of copper block: \(m_{Cu} = 2.0 \, kg\) - Initial temperature of copper block: \(T_{Cu(initial)} = 100.0^{\circ}C\) - Mass of water: \(m_{water} = 1.0 \, kg\) - Initial temperature of the water and the pot: \(T_{water(initial)} = T_{iron(initial)} = 25.0^{\circ}C\) - Mass of iron pot: \(m_{iron} = 2.0 \, kg\) - Specific heat capacities: \(c_{Cu} = 387 \, J/(kg \cdot K), c_{water} = 4186 \, J/(kg \cdot K), c_{iron} = 449 \, J/(kg \cdot K)\).
02

Apply the conservation of energy principle

As no energy is lost to the environment, the energy gained by water and the iron pot will be equal to the energy lost by the copper block. Therefore, we can write the conservation of energy as: $$m_{Cu}c_{Cu}(T_{Cu(initial)} - T_{final}) = m_{water}c_{water}(T_{final} - T_{water(initial)}) + m_{iron}c_{iron}(T_{final} - T_{iron(initial)})$$
03

Plug in the known values and solve for the final temperature

Substitute the known values into the equation: $$2.0 \, kg \cdot 387 \, J/(kg \cdot K)(100.0^{\circ}C - T_{final}) = 1.0 \, kg \cdot 4186 \, J/(kg \cdot K)(T_{final} - 25.0^{\circ}C) + 2.0 \, kg \cdot 449 \, J/(kg \cdot K)(T_{final} - 25.0^{\circ}C)$$ Now, simplify and solve for \(T_{final}\): $$774(100 - T_{final}) = 4186(T_{final} - 25) + 898(T_{final} - 25)$$ $$77400 - 774T_{final} = 4186T_{final} - 104650 + 898T_{final} - 22450$$ $$-774T_{final} + 4186T_{final} + 898T_{final} = -33100$$ $$4310T_{final} = 33100$$ $$T_{final} \approx 7.68^{\circ}C$$ The final temperature of the water (and the iron pot) will be approximately \(7.68^{\circ}C\).

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Most popular questions from this chapter

Consider the leaf of Problem \(70 .\) Assume that the top surface of the leaf absorbs \(70.0 \%\) of \(9.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2}\) of radiant energy, while the bottom surface absorbs all of the radiant energy incident on it due to its surroundings at \(25.0^{\circ} \mathrm{C} .\) (a) If the only method of heat loss for the leaf were thermal radiation, what would be the temperature of the leaf? (Assume that the leaf radiates like a blackbody.) (b) If the leaf is to remain at a temperature of \(25.0^{\circ} \mathrm{C}\) how much power per unit area must be lost by other methods such as transpiration (evaporative heat loss)?
You are given \(250 \mathrm{g}\) of coffee (same specific heat as water) at \(80.0^{\circ} \mathrm{C}\) (too hot to drink). In order to cool this to $\left.60.0^{\circ} \mathrm{C}, \text { how much ice (at } 0.0^{\circ} \mathrm{C}\right)$ must be added? Ignore heat content of the cup and heat exchanges with the surroundings.
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.
One end of a cylindrical iron rod of length \(1.00 \mathrm{m}\) and of radius \(1.30 \mathrm{cm}\) is placed in the blacksmith's fire and reaches a temperature of \(327^{\circ} \mathrm{C} .\) If the other end of the rod is being held in your hand \(\left(37^{\circ} \mathrm{C}\right),\) what is the rate of heat flow along the rod? The thermal conductivity of yal iron varies with temperature, but an average between the two temperatures is $67.5 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})$. (tutorial: conduction)
Small animals eat much more food per kg of body mass than do larger animals. The basal metabolic rate (BMR) is the minimal energy intake necessary to sustain life in a state of complete inactivity. The table lists the BMR, mass, and surface area for five animals. (a) Calculate the BMR/kg of body mass for each animal. Is it true that smaller animals must consume much more food per kg of body mass? (b) Calculate the BMR/m" of surface area. (c) Can you explain why the BMR/m \(^{2}\) is approximately the same for animals of different sizes? Consider what happens to the food energy metabolized by an animal in a resting state.
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