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 unit are you solving for? g °c cal none, i hate units

Explanation:

Step1: Identify the heat - capacity formula

The heat - capacity formula is $q = mc\Delta T$, 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 solve for $m$, so we can re - arrange the formula to $m=\frac{q}{c\Delta T}$.

Step2: Calculate the change in temperature

$\Delta T=T_2 - T_1=35.0^{\circ}C - 25.0^{\circ}C = 10.0^{\circ}C$.

Step3: Substitute the given values into the formula

We know that $q = 500$ cal, $c=0.0305$ cal/g°C, and $\Delta T = 10.0^{\circ}C$. Substituting these values into $m=\frac{q}{c\Delta T}$, we get $m=\frac{500\text{ cal}}{0.0305\text{ cal/g°C}\times10.0^{\circ}C}$.

Step4: Perform the calculation

$m=\frac{500}{0.0305\times10.0}=\frac{500}{0.305}\approx1639$ g.

The unit we are solving for is mass, and the unit of mass in the given context is grams (g).

Answer:

g