Example 17.2 Cooling a Hot Ingot Problem The temperature of a 0.0460 kg ingot of

Question

Example 17.2 Cooling a Hot Ingot

Problem The temperature of a 0.0460 kg ingot of metal is raised to 200.0°C and the ingot is then dropped into a light, insulated beaker containing 0.450 kg of water initially at 20.0°C. If the final equilibrium temperature of the mixed system is 22.4°C, find the specific heat of the metal.

Strategy Start off by using equation 17.4. By plugging equation 17.3 into each side you should then have an equality where you can substitute numerical values to solve for the unknown value of the specific heat.


Solution

Using equations 17.3 and 17.4, we can write a set of equations for solving this problem.

Qcold = -Qhot
mwcw(T - Tw) = -mxcx(T - Tx)
Now we'll substitute numerical values into the second equation.
(0.450 kg)(4186 J/kg·°C)(22.4°C - 20.0°C) = -(0.0460 kg)(cx)(22.4°C - 200.0°C)
After simplifying the equation we are able to solve for the specific heat cx.

cx = 553.3784 J/kg·°C


Using this specific heat we can compare it to known specific heats of different metals and take a guess as to which one this metal sample represents.




Exercise 17.2.1

What is the total energy transferred to the water in this example as the ingot is cooled?

Q = 4520.88 J


Exercise 17.2.2

A cowboy fires a silver bullet of mass 2.15 g with a muzzle speed of 260 m/s into a pine wall of a saloon, where the bullet stops. Assume that all the internal energy generated by the impact remains within the bullet. What is the temperature change of the bullet?

T = ____ °C

Explanation / Answer

Since you want only 17.2.2,
note specific heat of Ag is 0.233 J/g-K

Energy of bullet is ½mv2 = (0.5)(2.15 * 10-3 kg)(260 m/s)2 = 72.7 J

That is all converted to heat, and that much heat can raise the temperature of the bullet:

T = (72.7 J)/((0.233 J/g K)(2.15 g)) = 145 degrees C or K

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