Question

a 250 g rod of tantalum absorbed 10.1 kj and changed temperature from 25 °c to 250 °c. what is the specific heat of tantalum?
1.8×10⁻⁴ j/g°c
0.18 j/g°c
0.25 j/g°c
5.6 j/g°c

Explanation:

Step1: Convert energy unit

Convert 10.1 kJ to J. Since 1 kJ = 1000 J, then $Q=10.1\times1000\ J = 10100\ J$.

Step2: Calculate temperature - change

Calculate the change in temperature $\Delta T$. $\Delta T=T_2 - T_1$, where $T_1 = 25^{\circ}C$ and $T_2=250^{\circ}C$. So $\Delta T=250 - 25=225^{\circ}C$.

Step3: Use the heat - transfer formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat absorbed, $m$ is the mass, $c$ is the specific heat, and $\Delta T$ is the change in temperature. We want to find $c$, and we can re - arrange the formula to $c=\frac{Q}{m\Delta T}$. Given $m = 250\ g$, $Q = 10100\ J$, and $\Delta T=225^{\circ}C$. Then $c=\frac{10100\ J}{250\ g\times225^{\circ}C}\approx0.18\ J/g^{\circ}C$.

Answer:

0.18 J/g°C