Question

1. 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? heat = mass × specific heat × δt, where δt = t_final - t_initial 0.250 j/g°c 1.80 × 10⁻⁴ j/g°c 5.55 j/g°c 0.180 j/g°c

Explanation:

Step1: Convert heat to joules

$10.1\ \text{kJ} = 10.1 \times 1000 = 10100\ \text{J}$

Step2: Calculate temperature change

$\Delta t = 250^\circ\text{C} - 25^\circ\text{C} = 225^\circ\text{C}$

Step3: Rearrange formula for specific heat

$\text{specific heat} = \frac{\text{heat}}{\text{mass} \times \Delta t}$

Step4: Substitute values and solve

$\text{specific heat} = \frac{10100\ \text{J}}{250\ \text{g} \times 225^\circ\text{C}} = \frac{10100}{56250} \approx 0.180\ \text{J/g}^\circ\text{C}$

Answer:

0.180 J/g°C