Question

two objects are brought into contact. object #1 has mass 1.08 kg, specific heat capacity 0.76 j g⁻¹ °c⁻¹ and initial temperature 50.7 °c. object #2 has mass 50.1 g, specific heat capacity 4.14 j g⁻¹ °c⁻¹ and initial temperature 16.1 °c. what is the final temperature of the two masses after thermal equilibrium has been reached? assume the two objects are thermally isolated from everything else. express your answer in °c.

Explanation:

Step1: Convert mass of object 1 to grams

$m_1 = 1.08\ kg\times1000 = 1080\ g$

Step2: Set up heat - transfer equation

According to the principle of conservation of energy, $Q_1=-Q_2$. The heat - transfer formula is $Q = mc\Delta T$. So $m_1c_1(T - T_{1i})=-m_2c_2(T - T_{2i})$, where $m_1$ and $m_2$ are masses, $c_1$ and $c_2$ are specific heat capacities, $T_{1i}$ and $T_{2i}$ are initial temperatures, and $T$ is the final temperature.

Step3: Expand the equation

$m_1c_1T - m_1c_1T_{1i}=-m_2c_2T + m_2c_2T_{2i}$

Step4: Rearrange the equation to solve for T

$m_1c_1T+m_2c_2T=m_1c_1T_{1i}+m_2c_2T_{2i}$
$T=\frac{m_1c_1T_{1i}+m_2c_2T_{2i}}{m_1c_1 + m_2c_2}$

Step5: Substitute the given values

$m_1 = 1080\ g$, $c_1 = 0.76\ Jg^{-1}{^{\circ}C}^{-1}$, $T_{1i}=50.7^{\circ}C$, $m_2 = 50.1\ g$, $c_2 = 4.14\ Jg^{-1}{^{\circ}C}^{-1}$, $T_{2i}=16.1^{\circ}C$
$T=\frac{1080\times0.76\times50.7+50.1\times4.14\times16.1}{1080\times0.76 + 50.1\times4.14}$
$T=\frac{1080\times0.76\times50.7+50.1\times4.14\times16.1}{820.8+207.414}$
$T=\frac{41279.04+3341.5734}{1028.214}$
$T=\frac{44620.6134}{1028.214}\approx43.4^{\circ}C$

Answer:

$43.4^{\circ}C$