Question

1. 250.0 g of a metal releases 5000 j of energy and its temperature drops from 90.0°c to 15.70°c. what is the specific heat of the metal? (3pts)

heat = mass × specific heat × δt
where δt = t_final - t_initial
3.72 j/g°c
0.270 j/g°c
0.222 j/g°c
1.27 j/g
°c

Explanation:

Step1: Calculate temperature change

$\Delta t = t_{\text{final}} - t_{\text{initial}} = 15.7^\circ\text{C} - 90.0^\circ\text{C} = -74.3^\circ\text{C}$
(We use the absolute value for magnitude of heat change)

Step2: Rearrange heat formula for specific heat

Rearrange $q = m \times c \times \Delta t$ to solve for $c$:
$c = \frac{|q|}{m \times |\Delta t|}$

Step3: Substitute values into formula

$c = \frac{5000\ \text{J}}{250.0\ \text{g} \times 74.3^\circ\text{C}}$

Step4: Compute the result

$c = \frac{5000}{250.0 \times 74.3} = \frac{5000}{18575} \approx 0.270\ \text{J/g}^\circ\text{C}$

Answer:

$\square$ 0.270 J/g$^\circ$C