Question

Calculate $\mathrm{\Delta }E$ for each of the following. a. $q=-47\mathrm{kJ},w=+88\mathrm{kJ}$ b. $q=+82\mathrm{kJ},w=-47\mathrm{kJ}$ c. $q=+47\mathrm{kJ},w=0$ d. In which of these cases do the surroundings do work on the system?

Short answer

The change in internal energy ($\mathrm{\Delta }E$) for each case is calculated as follows: a. $\mathrm{\Delta }E=41\mathrm{kJ}$ b. $\mathrm{\Delta }E=35\mathrm{kJ}$ c. $\mathrm{\Delta }E=47\mathrm{kJ}$ The surroundings do work on the system only in case a.

Step-by-step solution

Case a: Calculate $\mathrm{\Delta }E$

To calculate the change in internal energy for case a, we will plug in the given values of $q$ and $w$ into the formula: $$\mathrm{\Delta }E=(-47\mathrm{kJ})+(+88\mathrm{kJ})$$ Now, we simply add the two values together: $$\mathrm{\Delta }E=41\mathrm{kJ}$$

Case b: Calculate $\mathrm{\Delta }E$

For case b, plug in the given values of $q$ and $w$ into the formula: $$\mathrm{\Delta }E=(+82\mathrm{kJ})+(-47\mathrm{kJ})$$ Now, add the two values together: $$\mathrm{\Delta }E=35\mathrm{kJ}$$

Case c: Calculate $\mathrm{\Delta }E$

For case c, plug in the given values of $q$ and $w$ into the formula: $$\mathrm{\Delta }E=(+47\mathrm{kJ})+(0)$$ Since work done ($w$) is zero, the change in internal energy is equal to the heat added to the system: $$\mathrm{\Delta }E=47\mathrm{kJ}$$

Identify cases where surroundings do work on the system

The surroundings do work on the system when the work ($w$) is positive. Looking at the given values for each case, we find that: - In case a, $w=+88$ kJ (positive) - In case b, $w=-47$ kJ (negative) - In case c, $w=0$ kJ (zero) Thus, the surroundings do work on the system only in case a.