# "2659 J" of heat energy are required to heat a substance with a mass of "24.25 g" from "23.7"^@"C" to "163.5"^@"C". What is its specific heat capacity?

Nov 16, 2016

The specific heat capacity is 784.3 J/(C·kg).

#### Explanation:

So, the equation we are using is $Q = m c \Delta T$.

=> Where $Q$ is the amount of thermal energy in joules, $J$.
=> Where $m$ is the mass in kilograms, $k g$.
=> Where $c$ is the specific heat capacity.
=> Where $\Delta T$ is the change in temperature in degrees Celsius.

$24.25 g = 0.02425 k g$

$Q = m c \Delta T$

$Q = m c \left({T}_{2} - {T}_{1}\right)$

$\text{2659 J" = "0.02425 kg"xxcxx"(163.5 - 23.7) °C}$

$\text{2659 J" = "0.02425 kg"xxcxx"139.8 °C}$

$\text{19.02 J/°C"= "0.02425 kg} \times c$

$c = \text{784.3 J/(°C·kg)}$

The specific heat capacity is 784.3 J/(°C·kg).

Hope this helps :)

Nov 28, 2016

The specific heat is 0.7843"J"/("g"*^@"C").

#### Explanation:

Apply the equation $Q = m c \Delta T$, where $Q$ is the energy in Joules, $m$ is the mass in grams, $c$ is the specific heat, and $\Delta T$ is change in temperature. $\Delta T = {T}_{\text{final"-T_"initial}}$

Known
$Q = \text{2659 J}$
$m = \text{24.25 g}$
${T}_{\text{initial"="23.7"^@"C}}$
${T}_{\text{final"="163.5"^@"C}}$
$\Delta T = \text{163.5"^@"C"-"23.7"^@"C"="139.8"^@"C}$

Unknown
$c$

Solution
Rearrange the equation $Q = m c \Delta T$ to isolate $c$. Substitute the known values into the equation and solve.

$c = \frac{Q}{m \cdot \Delta T}$

$c = \left(2659 \text{J")/(24.25"g"*139.8^@"C}\right)$

c=0.7843"J"/("g"*^@"C")