Question

how many calories would be gained when 1.0000 x 10^6 g of gold (one of my favorite elements, although i like most of them) is heated from 80.0 to 91.0 °c? the specific heat of au is 0.03050 cal/g°c.
the conversion factor is 1 since this will give us calories as the answer.
we must then cancel out the other units.
we multiply by 1.0000 x 10^6 g and then multiply by 2 to get the final answer of 3.
1 g 1 °c 0.03050 cal
a. ----------- b. ----------- c. 1.0000 x 10^6 g d. 0.0305 cal/g°c
0.03050 cal 1 g 1 °c
e. 91.0 °c f. 11 °c g. 11.0 °c h. 3.355 x 10^5 cal i. 3.4 x 10^5 cal
j. 3.36 x 10^5 cal

Explanation:

Step1: Calculate temperature change

$\Delta T=91.0 - 80.0=11.0^{\circ}C$

Step2: Identify conversion factor

The specific - heat formula is $q = mc\Delta T$, where $q$ is heat, $m$ is mass, $c$ is specific heat, and $\Delta T$ is temperature change. The conversion factor to get calories is the specific heat $c = 0.03050\ cal/g^{\circ}C$, so 1 is D.

Step3: Calculate heat gained

$m = 1.0000\times10^{6}\ g$, $c = 0.03050\ cal/g^{\circ}C$, $\Delta T=11.0^{\circ}C$
$q=(1.0000\times 10^{6}\ g)\times(0.03050\ cal/g^{\circ}C)\times(11.0^{\circ}C)=3.355\times 10^{5}\ cal$, so 3 is H. And 2 is the temperature - change value which is G.

Answer:

1. D. $0.0305\ cal/g^{\circ}C$
2. G. $11.0^{\circ}C$
3. H. $3.355\times 10^{5}\ cal$