# How much heat is absorbed by a 2000 kg granite boulder as energy from the sun causes its temperature to change from 10°C to 29°C?

Apr 24, 2017

3.002 MJ

#### Explanation:

Heat absorbed,
$H = m . c . \partial \left(\theta\right)$
Where, m= mass,
C= specific heat capacity. Assume $c = 790 \frac{J}{k g} / K$ for granite.
$\partial \left(\theta\right)$ is the change in temperature.
Hence, $\partial \left(\theta\right) = 29 - 10 = 19 K$

(Note: For difference of temperature, Celsius degrees and Kelvin are interchangeable, as the interval for both scales is the same.)

Then we have,
$H = 2000 \cdot 790 \cdot 19 = 3002000 J = 3.002 M J$

Apr 24, 2017

Apply the equation

$\textcolor{red}{Q = m C p \Delta T}$

$\textcolor{b l u e}{\text{Where Q is heat absorbed}}$

$\textcolor{b l u e}{\text{Cp is the specific heat}}$

color(blue)(DeltaT "is " t_"final" - t_"initial"

$\textcolor{b l u e}{\text{m is mass in grams}}$

The $C p$ of granite

$C p$ of granite is $\text{0.79J/g} \cdot {C}^{\circ}$

Convert 2000kg to grams = $2 \cdot {10}^{6} g$

$2 \cdot {10}^{6} g \times \text{0.79J/g} \cdot {C}^{\circ} \times \left(29 {C}^{\circ} - 10 {C}^{\circ}\right) = 30 020 000 J$