# Question eca79

Nov 8, 2015

$\text{60 K}$

#### Explanation:

In order to be able to solve this problem, you need to know the specific heat of silver, which is listed as being equal to

c_"silver" = 0.23"J"/("g K")

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

Now, the idea here is that a substance's specific heat tells you how much energy must be provided in order to increase the temperature of $\text{1 g}$ of that substance by $\text{1 K}$.

In your case, you know that you provide a $\text{32-g}$ sample of silver with $\text{300 J}$ worth of heat and want to determine how much will the sample's temperature increase.

The equation that establishes a relationship between heat absorbed and change in temperature looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat absorbed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the final temperature minus the initial temperature of the sample.

So, you need to rearrange this equation and solve for $\Delta T$

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

Plug in your values to get

DeltaT = (300color(red)(cancel(color(black)("J"))))/(32color(red)(cancel(color(black)("g"))) * 0.23color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) * "K")) = "40.8 K"

This tells you that the temperature of the sample changed by $\text{40.8 K}$. Therefore, the final temperature will be

$\Delta T = {T}_{\text{final" - T_"initial}}$

${T}_{\text{final" = "40.8 K" + "20 K" = "60.8 K}}$

You need to round this off to one sig fig, the number of sig figs you have for the heat absorbed and for the initial temperature

T_"final" = color(green)("60 K")#