Solution Found!
In are given the equation used to solve a
Chapter 17, Problem 75P(choose chapter or problem)
In Problems 74 through 76 you are given the equation used to solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation.
b. Finish the solution of the problem.
\(\left(200 \times 10^{-6} \mathrm{~m}^{3}\right)\left(13,600 \mathrm{~kg} / \mathrm{m}^{3}\right)\)
\(\times(140 \mathrm{~J} / \mathrm{kgK})\left(90^{\circ} \mathrm{C}-15^{\circ} \mathrm{C}\right)+(0.50 \mathrm{~kg})(449 \mathrm{~J} / \mathrm{kg} K)\left(90^{\circ} \mathrm{C}-T_{i}\right)=0\)
Equation Transcription:
Text Transcription:
(200x10^-6m3)(13,600 kg/m^3)
x(140J/kg K)90^circ C-15^circC)+(0.50 kg)(449J/kg K)(90^circC-T_i)=0
Questions & Answers
QUESTION:
In Problems 74 through 76 you are given the equation used to solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation.
b. Finish the solution of the problem.
\(\left(200 \times 10^{-6} \mathrm{~m}^{3}\right)\left(13,600 \mathrm{~kg} / \mathrm{m}^{3}\right)\)
\(\times(140 \mathrm{~J} / \mathrm{kgK})\left(90^{\circ} \mathrm{C}-15^{\circ} \mathrm{C}\right)+(0.50 \mathrm{~kg})(449 \mathrm{~J} / \mathrm{kg} K)\left(90^{\circ} \mathrm{C}-T_{i}\right)=0\)
Equation Transcription:
Text Transcription:
(200x10^-6m3)(13,600 kg/m^3)
x(140J/kg K)90^circ C-15^circC)+(0.50 kg)(449J/kg K)(90^circC-T_i)=0
ANSWER:
Step 1 of 2
Part a
We are required to state a realistic problem based on the given equation.
The density of mercury is 13,600 . From table 17.2, the specific heat of mercury is 140 J/kgK. Similarly, the specific heat of iron is 449 J/kgK. Therefore, the metals involved are iron and mercury.
The problem
An iron ball of mass 0.50 kg is dropped into a container having mercury at 15 °C and the final temperature of mercury becomes 90 °C. Calculate the initial temperature of the iron ball.