# Question #89fac

Apr 16, 2016

${c}_{\text{B" = 2 xx c_"A}}$

#### Explanation:

You can answer this question without actually calculating the specific heats of the two substances.

As you know, a substance's specific heat tell you how much heat is needed in order to increase the temperature of $\text{1 g}$ of this substance by ${1}^{\circ} \text{C}$.

Notice that you're dealing with equal masses of substance $\text{A}$ and substance $\text{B}$ and that the increase in temperature is the same for both substances.

This means that the difference between the amount of heat needed to cause the same change in temperature for equal masses of both substances will reflect the difference in their specific heats.

More specifically, the substance that requires more heat to register the same change in temperature will have a higher specific heat.

In this case, substance $\text{B}$ requires two times more heat than substance $\text{A}$, so its specific heat must be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{\text{B" = 2 xx c_"A}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

To prove this result, you can calculate the two specific heats by using the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat gained
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

Convert the masses from kilograms to grams first

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{1 kg" = 10^3"g}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Rearrange the equation to solve for $c$

$q = m \cdot c \cdot \Delta T \implies c = \frac{q}{m \cdot \Delta T}$

Plug in your values to get

${c}_{\text{A" = "500 J"/(5 * 10^3"g" * 2^@"C") = "0.05 J g"^(-1)""^@"C}}^{- 1}$

${c}_{\text{B" = "1000 J"/(5 * 10^3"g" * 2^@"C") = "0.1 J g"^(-1)""^@"C}}^{- 1}$

As predicted, the specific heat of substance $\text{B}$ is twice as high as the specific heat of substance $\text{A}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{c}_{\text{B" = 2 xx c_"A}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$