# How many grams of silver are needed to absorb 1.00 kJ of energy changing the temperature from 33.0°C to 41.8°C?

Apr 15, 2017

$484 g$

#### Explanation:

To solve this, we use the formula for heat absorption, $Q = m T c$, where
$Q$ is the heat energy absorbed, in Joules,
$m$ is the mass of the object usually in grams,
$T$ is the temperature change of the object, in Celcius or Kelvin, and
$c$ is the specific heat capacity of the material, in joules per gram °C. This is the amount of energy it takes for one gram of this substance to heat up by 1K.

We then rearrange this equation making $m$ the subject:
$m = \frac{Q}{T c}$

The question gives us the values $Q = 1 k J = 1000 J$, and we can calculate $T$ by using

$T = \text{Final Temperature" - "Initial Temperature" = 41.8"°C" - 33.0"°C}$
$T = 8.80 \text{°C = 8.80K}$

Now the only remaining value we need is $c$. Since this depends on the material involved, we have to look up the value for silver.
I found 3 sources: hyperphysics, periodictable.com, and ptable.com.

These put the value at $0.233 J {g}^{-} 1 {K}^{-} 1$, $0.235 J {g}^{-} 1 {K}^{-} 1$, and $0.235 J {g}^{-} 1 {K}^{-} 1$ respectively. I will use $0.235 J {g}^{-} 1 {K}^{-} 1$ here.

Now we substitute into the equation:

$m = \frac{Q}{T c} = \frac{1000 J}{8.80 K \times 0.235 J {g}^{-} 1 {K}^{-} 1} = 484 g \text{ (3 s.f.)}$

If you have been given a value for the specific heat capacity of silver which is not $0.235 J {g}^{-} 1 {K}^{-} 1$, then your final answer may be slightly different.