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 calories? give two answers select 2 correct answer(s) 1 g °c 500. cal (-----------)(-----------) 0.0305 cal 1 0.0305 cal 1 (-----------)(-----------) 1 g °c 500. cal (1 g °c/0.0305 cal) (500. cal)

Explanation:

Step1: Recall heat - mass - specific heat formula

The formula for heat $Q = mc\Delta T$, where $Q$ is heat, $m$ is mass, $c$ is specific heat, and $\Delta T$ is change in temperature. We want to solve for $m$, so $m=\frac{Q}{c\Delta T}$. Here, $Q = 500$ cal, $c=0.0305$ cal/g°C, and $\Delta T=(35 - 25)=10$°C. To cancel out calories in the calculation of mass, we use the ratio of units such that the calorie units in the numerator and denominator cancel.

Step2: Analyze unit - cancellation for mass calculation

We know that to get mass in grams, we need to cancel out the calorie unit. If we have a fraction with calories in the denominator and the product of grams and degrees - Celsius in the numerator, when multiplied by the quantity of heat (in calories), the calorie units will cancel. The correct combinations for canceling out calories are when we multiply the heat value (500 cal) by $\frac{1\ g^{\circ}C}{0.0305\ cal}$.

Answer:

$(1\ g^{\circ}C/0.0305\ cal)(500\ cal)$ is the correct way to cancel out calories to find the mass of the gold sample. So the correct option is the one with $(1\ g^{\circ}C/0.0305\ cal)(500\ cal)$.