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 we start the problem? give two correct answers select 2 correct answer(s) 1 g °c (------------) 0.0305 cal 0.0305 cal (------------) 1 g °c (0.0305 cal/1 g °c)

Explanation:

Step1: Recall the heat - transfer formula

The heat - transfer 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 find $m$, so we can re - arrange the formula to $m=\frac{q}{c\Delta T}$. The specific heat $c = 0.0305\ cal/g^{\circ}C$ means that it takes $0.0305$ calories of heat to raise the temperature of $1\ g$ of the substance by $1^{\circ}C$. In other words, the ratio of heat to mass and temperature change is $\frac{0.0305\ cal}{1\ g^{\circ}C}$.

Step2: Identify the correct starting points

To start the problem using the re - arranged formula $m=\frac{q}{c\Delta T}$, we need to use the relationship between the units in the specific - heat value. The two correct ways to start are considering the ratio of the specific heat. The specific heat $c = 0.0305\ cal/g^{\circ}C$ can be written as a fraction $\frac{0.0305\ cal}{1\ g^{\circ}C}$ or its reciprocal $\frac{1\ g^{\circ}C}{0.0305\ cal}$.

Answer:

1. $\frac{0.0305\ cal}{1\ g^{\circ}C}$
2. $\frac{1\ g^{\circ}C}{0.0305\ cal}$