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Chapter 7: Problem 8

Brass has a density of \(8.40 \mathrm{g} / \mathrm{cm}^{3}\) and a specific heat of \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} . \mathrm{A} 15.2 \mathrm{cm}^{3}\) piece of brass at an initial temperature of \(163^{\circ} \mathrm{C}\) is dropped into an insulated container with \(150.0 \mathrm{g}\) water initially at \(22.4^{\circ} \mathrm{C}\) What will be the final temperature of the brass-water mixture?

Short Answer

Expert verified
The final temperature of the brass-water mixture is reached by equating the heat lost by the brass to the heat gained by the water and solving for the temperature. The exact value needs calculations based on the formulated equation and provided parameters.

Step by step solution

01

Identify given parameters

We are given the following parameters: brass density is \(8.40 \mathrm{g} / \mathrm{cm}^{3}\), brass specific heat is \(0.385 \mathrm{Jg}^{-1} \mathrm{C}^{-1}\), volume of the brass piece is \(15.2 \mathrm{cm}^{3}\), its initial temperature is \(163^{\circ}\mathrm{C}\), the mass of water is \(150.0 \mathrm{g}\), and its initial temperature is \(22.4^{\circ} \mathrm{C}\)
02

Calculate the loss of heat from the brass

First, find the mass of the brass using the formula \( \text{density} = \text{mass} / \text{volume} \). So, \( \text{mass} = \text{density} \times \text{volume} = 8.40 \mathrm{g} / \mathrm{cm}^{3} \times 15.2 \mathrm{cm}^{3}\) then using the formula for heat exchange \( Q = mcΔT \), we can calculate the heat loss of the brass. Here, \(m\) is mass, \(c\) is specific heat and \(ΔT\) is change in temperature, which is \(163 - T_{final}\)
03

Calculate the gain of heat by water

Applying the same formula \( Q = mcΔT \) on water, its heat gain is \(150.0 \mathrm{g} \times 4.18 \mathrm{Jg}^{-1} \mathrm{C}^{-1} \times (T_{final}-22.4^{\circ}\mathrm{C})\). Note, we've used the specific heat of water, which is \(4.18 \mathrm{Jg}^{-1} \mathrm{C}^{-1}\)
04

Solve for final temperature

Since the heat gained by the water is equal to the heat lost by the brass, we can equate the two equations from step 2 and 3 and solve for \(T_{final}\)

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Most popular questions from this chapter

\(\Delta U=100 \mathrm{J}\) for a system that gives off \(100 \mathrm{J}\) of heat and (a) does no work; (b) does 200 J of work; (c) has 100 J of work done on it; (d) has 200 J of work done on it.

What is the change in internal energy of a system if the surroundings (a) transfer 235 J of heat and 128 J of work to the system; (b) absorb 145 J of heat from the system while doing \(98 \mathrm{J}\) of work on the system; (c) exchange no heat, but receive 1.07 kJ of work from the system?

Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction.Upon complete combustion, the indicated substances evolve the given quantities of heat. Write a balanced equation for the combustion of \(1.00 \mathrm{mol}\) of each substance, including the enthalpy change, \(\Delta H\), for the reaction. (a) \(0.584 \mathrm{g}\) of propane, \(\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{g}),\) yields \(29.4 \mathrm{kJ}\) (b) \(0.136 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O}(\mathrm{s}),\) yields \(5.27 \mathrm{kJ}\) (c) \(2.35 \mathrm{mL}\) of acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(\mathrm{l})(d=0.791\) \(\mathrm{g} / \mathrm{mL}),\) yields \(58.3 \mathrm{kJ}\)

The following substances undergo complete combustion in a bomb calorimeter. The calorimeter assembly has a heat capacity of \(5.136 \mathrm{kJ} /^{\circ} \mathrm{C} .\) In each case, what is the final temperature if the initial water temperature is \(22.43^{\circ} \mathrm{C} ?\) \(\begin{array}{lllll}\text { (a) } 0.3268 & \text { g caffeine, } & \mathrm{C}_{8} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{N}_{4} & \text { (heat of }\end{array}\) combustion \(=-1014.2 \mathrm{kcal} / \mathrm{mol} \text { caffeine })\) (b) \(1.35 \mathrm{mL}\) of methyl ethyl ketone, \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}(1)\) \(d=0.805 \mathrm{g} / \mathrm{mL}\) (heat of combustion \(=-2444 \mathrm{kJ} / \mathrm{mol}\) methyl ethyl ketone).

A 465 g chunk of iron is removed from an oven and plunged into \(375 \mathrm{g}\) water in an insulated container. The temperature of the water increases from 26 to \(87^{\circ} \mathrm{C}\). If the specific heat of iron is \(0.449 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1},\) what must have been the original temperature of the iron?
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