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? what is the final answer? give two correct answers select 2 correct answer(s) 1 g °c 500. cal 1 (----------------)(----------------)(----------------) = 1.64×10³ g 0.0305 cal 1 10.0 °c 0.0305 cal 1 10.0 °c (----------------)(----------------)(----------------) = 6.10×10⁻⁴ g 1 g °c 500. cal 1 (0.0305 cal/1 g °c) (1/500. cal) (10.0 °c) = 6.10×10⁻⁴ g

Explanation:

Step1: Calculate temperature change

$\Delta T=35.0 - 25.0=10.0^{\circ}C$

Step2: Use heat - mass - specific heat formula

The formula for heat is $Q = mc\Delta T$, where $Q$ is heat, $m$ is mass, $c$ is specific heat and $\Delta T$ is temperature change. We want to find $m$, so we can re - arrange the formula to $m=\frac{Q}{c\Delta T}$.

Step3: Substitute values

Given $Q = 500$ cal, $c=0.0305$ cal/g°C and $\Delta T = 10.0^{\circ}C$. Then $m=\frac{500\text{ cal}}{0.0305\text{ cal/g°C}\times10.0^{\circ}C}=\frac{500}{0.305}\text{ g}\approx1.64\times 10^{3}\text{ g}$

Answer:

1. First correct answer: $\frac{1\text{ g}^{\circ}C}{0.0305\text{ cal}}\times500\text{ cal}\times\frac{1}{10.0^{\circ}C}=1.64\times 10^{3}\text{ g}$
2. Second correct answer: None of the other options are correct as the mass calculation based on the heat - capacity formula gives $m = 1.64\times 10^{3}\text{ g}$ and the other expressions do not yield the correct result.