Question

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 2.30 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than 10.0 °C? Use Cm = 30.8 J/(K·mol) for Na(l) at 500 K.

Answer 1

Answer on Question #55464 - Chemistry - General chemistry

Question:

Liquid sodium is being considered as an engine coolant. How many grams of liquid sodium (minimum) are needed to absorb 2.30 MJ of energy (in the form of heat) if the temperature of the sodium is not to increase by more than $10.0^{\circ}\mathrm{C}$? Use $\mathrm{Cm} = 30.8\mathrm{J / (K\cdot mol)}$ for Na(I) at $500\mathrm{K}$.

Solution:

The amount of energy absorbed can be calculated using the heat capacity:

$$Q = m c (T _ {2} - T _ {1})$$

where $m$ is the mass of sodium, $c$ heat capacity, $T_{2}$ is the final temperature (510 K in our case) and $T_{1}$ is the initial temperature (500 K).

Then, the expression for the mass will be:

$$m = \frac {Q}{c \left(T _ {2} - T _ {1}\right)}$$

In our case, the heat capacity is given per mole of sodium, so we should divide it by the molar mass of sodium:

$$c = \frac {c _ {M}}{M} = \frac {c _ {M}}{2 3}$$

Then:

$$m = \frac {Q}{c (T _ {2} - T _ {1})} = \frac {2 . 3 0 \times 1 0 ^ {6} \times 2 3}{3 0 . 8 \times (1 0)} = 1. 7 2 \times 1 0 ^ {5} g,$$

So the mass of liquid sodium should be $1.72 \times 10^{5} \, \text{g}$, or $172 \, \text{kg}$.

Answer: $172\mathrm{kg}$

www.AssignmentExpert.com