Question

the sp ht$_{au}$ = 0.03050 cal/g$^{circ}$c. a sample of au lost 2.5000 x 10$^{4}$ kcal when cooled from 125.0 to 100.0 $^{circ}$c. what is the mass of the sample? the conversion factor is 1 since that gives us the final unit of grams. we need to cancel out all other units. first we multiply by 2, then we divide by 3. the final answer is 4. 1 g 1 $^{circ}$c 0.0305 cal a. ---------- b. ---------- c. 2.5000 x 10$^{4}$ kcal 0.0305 cal 1 g 1 $^{circ}$c d. 2.5000 x 10$^{7}$ cal e. 125.0 $^{circ}$c f. 25 $^{circ}$c g. 25.0 $^{circ}$c h. 3.28 x 10$^{7}$ g i. 3.05 x 10$^{4}$ g j. 3.05 x 10$^{8}$ g

Explanation:

Step1: Convert kcal to cal

We know that $1\ kcal = 1000\ cal$, so $2.5000\times 10^{4}\ kcal=2.5000\times 10^{4}\times1000 = 2.5000\times 10^{7}\ cal$.

Step2: Calculate the temperature - change

The initial temperature $T_1 = 125.0^{\circ}C$ and the final temperature $T_2 = 100.0^{\circ}C$. The temperature - change $\Delta T=T_1 - T_2=125.0 - 100.0=25.0^{\circ}C$.

Step3: Use the heat - capacity formula $Q = mc\Delta T$ to find mass $m$

The specific heat capacity $c = 0.0305\ cal/g^{\circ}C$, and $Q = 2.5000\times 10^{7}\ cal$, $\Delta T = 25.0^{\circ}C$. Rearranging the formula $m=\frac{Q}{c\Delta T}$.
Substitute the values: $m=\frac{2.5000\times 10^{7}\ cal}{0.0305\ cal/g^{\circ}C\times25.0^{\circ}C}$.
First, multiply the denominator: $0.0305\ cal/g^{\circ}C\times25.0^{\circ}C = 0.7625\ cal/g$.
Then, $m=\frac{2.5000\times 10^{7}\ cal}{0.7625\ cal/g}\approx3.28\times 10^{7}\ g$.

We use the conversion factor $\frac{1\ g}{0.0305\ cal\times1^{\circ}C}$ to cancel out the units of cal and $^{\circ}C$ and get the mass in grams. First, we multiply by the amount of heat in cal ($2.5000\times 10^{7}\ cal$), then we divide by the product of specific - heat and temperature - change ($0.0305\ cal\times25.0^{\circ}C$).

Answer:

1. $\frac{1\ g}{0.0305\ cal\times1^{\circ}C}$
2. $2.5000\times 10^{7}\ cal$
3. $0.0305\ cal\times25.0^{\circ}C$
4. $3.28\times 10^{7}\ g$