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.

75. $\left(200\times {10}^{-6}{\mathrm{m}}^3\right)\left(13,600\mathrm{kg}/{\mathrm{m}}^3\right)\times (140\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-15°\mathrm{C}\right)$

$+(0.50\mathrm{kg})(449\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-T_{\mathrm{i}}\right)=0$

Short answer

a) We are to find the initial temperature of the iron.

b) The solution of the problem is$-{56}^{\circ }C.$

Step-by-step solution

Given Information (Part a)

Equation:$\left(200\times {10}^{-6}{\mathrm{m}}^3\right)\left(13,600\mathrm{kg}/{\mathrm{m}}^3\right)\times (140\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-15°\mathrm{C}\right)$

$+(0.50\mathrm{kg})(449\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-T_{\mathrm{i}}\right)=0$

Explanation (Part a)

(a) As we can see, this is a calorimetry problem.

We are given that $200{\mathrm{cm}}^3$of a substance of density $13600\mathrm{kg}/{\mathrm{m}}^3$(which means it is mercury), with a heat capacity of $140\mathrm{J}/\mathrm{molK}$, was at an initial temperature of $15°\mathrm{C}$and when a mass of $0.5\mathrm{kg}$of a substance of heat capacity $449\mathrm{J}/\mathrm{kgK}$(which means it is iron) at an initial temperature $T_i$was dropped into it, it reached an equilibrium temperature of $90°\mathrm{C}$.We are to find the initial temperature of the iron.

Final Answer (Part a)

The realistic problem is:

We are given that $200{\mathrm{cm}}^3$of a substance of density $13600\mathrm{kg}/{\mathrm{m}}^3$, with a heat capacity of $140J/molK$, was at an initial temperature of $15°\mathrm{C}$and when a mass of $0.5\mathrm{kg}$of a substance of heat capacity $449\mathrm{J}/\mathrm{kgK}$at an initial temperature $T_j$was dropped into it, it reached an equilibrium temperature of$90°\mathrm{C}$ We are to find the initial temperature of the iron.

Given Information (Part a)

Equation: $\left(200\times {10}^{-6}{\mathrm{m}}^3\right)\left(13,600\mathrm{kg}/{\mathrm{m}}^3\right)\times (140\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-15°\mathrm{C}\right)$

$+(0.50\mathrm{kg})(449\mathrm{J}/\mathrm{kgK})\left(90°\mathrm{C}-T_{\mathrm{i}}\right)=0$

Explanation (Part b)

(b) Again, we would have normally solved this problem parametrically, but since we are given a numerically-substituted expression, let's continue with it.

It is clear, by the symbol and the fact that it must have the same unit as the other part of the multiplying factor, that the quantity we are looking for is temperature.

We can put one factor to the other sign (notice that we put no minus in front of the entire expression, since we multiplied the factor of $\left(90-T_i\right)$with minus one) and find

$2·{10}^{-4}{\mathrm{m}}^3·13600\mathrm{kg}/{\mathrm{m}}^3·140\mathrm{J}/\mathrm{kgK}·75°\mathrm{C}=0.5\mathrm{kg}·449\mathrm{J}/\mathrm{kgK}·\left(T_i-90{}^{\circ }\mathrm{C}\right)$

We can simplify the units of volume, mass and temperature on each of the sides, and the units of energy across the two sides.

(This, strictly speaking, means that our result will be dimensionless ; or, we'd have to define a new variable with units $1/K)$

Thus we have

$2·{10}^{-4}·13600·140·75=0.5·449·T_i-0.5·449·90$

We can therefore finally find our temperature as

$T_i=\frac{28560+20205}{224.5}=\underset{¯}{217\mathrm{K}}$

In degrees Celsius, our result would be

$T_i=-56°\mathrm{C}$

Final Answer (Part b)

The solution of the problem is$-{56}^{\circ }C.$