Question

after transfer of 1.50 kj of thermal energy to a 0.505 - kg block of copper the temperature is 45.4 °c. the specific heat capacity of copper is 0.385 j g⁻¹ °c⁻¹. calculate the initial temperature of the copper.

Explanation:

Step1: Recall the heat - transfer formula

$Q = mc\Delta T$, where $Q$ is the heat energy transferred, $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T=T_{final}-T_{initial}$. We want to find $T_{initial}$, so we can re - arrange the formula to $T_{initial}=T_{final}-\frac{Q}{mc}$.

Step2: Convert the mass and heat energy to SI units

The mass $m = 0.505\ kg$, the heat energy $Q=1.50\ kJ = 1500\ J$, and the specific heat capacity $c = 0.385\ Jg^{-1}{\hspace{0.1cm}}^{\circ}C^{-1}=385\ Jkg^{-1}{\hspace{0.1cm}}^{\circ}C^{-1}$, and $T_{final}=45.4^{\circ}C$.

Step3: Substitute the values into the re - arranged formula

$T_{initial}=45.4-\frac{1500}{0.505\times385}$
First, calculate the denominator: $0.505\times385 = 0.505\times385=194.425$.
Then, calculate $\frac{1500}{194.425}\approx7.725$.
Finally, $T_{initial}=45.4 - 7.725=37.7^{\circ}C$

Answer:

$37.7$