Question

it required 500. cal to heat a sample of gold from 25.0 to 35.0 °c. what is the mass if the specific heat is 0.0305 cal/g°c? how do you cancel out °c? give two correct answers select 2 correct answer(s) 1 g °c 500. cal 1 (-----------)(-----------)(-----------) 0.0305 cal 1 10.0 °c 0.0305 cal 1 10.0 °c (-----------)(-----------)(-----------) 1 g °c 500. cal 1 (0.0305 cal/1 g °c) (1/500. cal) (10.0 °c)

Explanation:

Step1: Recall the heat - mass - specific heat formula

The formula for heat $q = mc\Delta T$, where $q$ is the heat energy, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature. We want to cancel out the units to get the mass in grams. The unit of specific heat $c$ is $\text{cal/g}^ {\circ}\text{C}$, $q$ is in cal and $\Delta T$ is in $^{\circ}\text{C}$. To cancel out $^{\circ}\text{C}$, we need to multiply by a factor that has $^{\circ}\text{C}$ in the denominator.

Step2: Analyze the units for cancellation

We know that $m=\frac{q}{c\Delta T}$. The units of $q$ is cal, $c$ is $\text{cal/g}^ {\circ}\text{C}$ and $\Delta T$ is $^{\circ}\text{C}$. If we multiply $\frac{1}{c}$ (which has units $\text{g}^ {\circ}\text{C/cal}$) by $q$ (cal) and $\frac{1}{\Delta T}$ (with units $1/^{\circ}\text{C}$), the $^{\circ}\text{C}$ and cal units will cancel out.
The correct combinations for canceling out $^{\circ}\text{C}$ are:

1. $\frac{1\ g^{\circ}\text{C}}{0.0305\ cal}\times\frac{500\ cal}{1}\times\frac{1}{10.0^{\circ}\text{C}}$
2. $\frac{1}{0.0305\ cal/1\ g^{\circ}\text{C}}\times\frac{1}{500\ cal}\times(10.0^{\circ}\text{C})$

Answer:

The two correct answers are:

1. $\frac{1\ g^{\circ}\text{C}}{0.0305\ cal}\times\frac{500\ cal}{1}\times\frac{1}{10.0^{\circ}\text{C}}$
2. $\frac{1}{0.0305\ cal/1\ g^{\circ}\text{C}}\times\frac{1}{500\ cal}\times(10.0^{\circ}\text{C})$