# How would you calculate the specific beat of silver, given that 280 g of the metal absorbed 136 calories when changed from 22.13°C to 30.79°C?

Dec 1, 2016

c_"Ag"=0.056"cal"/("g"*^@"C")

#### Explanation:

The equation you'll need is the following:

where $Q$ is in calories, $c$ is in "cal"/("g"*^@"C"), $m$ is mass in grams, and $\Delta T$ is the change in Celcius temperature. $\Delta T = {T}_{\text{final"-T_"initial}}$

$Q = \text{136 cal}$
$m = \text{280 g}$
${T}_{\text{initial"="22.13"^@"C}}$
${T}_{\text{final"="30.79"^@"C}}$
$\Delta T = \text{30.79"^@"C"-"22.13"^@"C}$"="8.66"^@"C"

Rearrange the equation to isolate $c$.

${c}_{\text{Ag}} = \frac{Q}{m \Delta T}$

c_"Ag"=(136"cal")/(280"g"*8.66^@"C")

c_"Ag"=0.056"cal"/("g"*^@"C")#