Question

1. 45.0 g of water is at 20.0°c. the water releases 2000 j of energy as heat. what (1pts) is the new temperature of the water in kelvin? heat = mass × specific heat × δt where δt = t_final - t_initial □ 19.9 k □ 30.6 k □ -168 k □ 282 k

Explanation:

Step1: Convert initial temp to Kelvin

$T_{initial} = 20.0^\circ\text{C} + 273.15 = 293.15\ \text{K}$

Step2: Rearrange heat formula for $\Delta T$

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

Step3: Substitute known values

Specific heat of water is $4.184\ \frac{\text{J}}{\text{g·K}}$, heat released = $-2000\ \text{J}$
$\Delta T = \frac{-2000\ \text{J}}{45.0\ \text{g} \times 4.184\ \frac{\text{J}}{\text{g·K}}} \approx -10.6\ \text{K}$

Step4: Calculate final temperature

$T_{final} = T_{initial} + \Delta T$
$T_{final} = 293.15\ \text{K} - 10.6\ \text{K} \approx 282.6\ \text{K} \approx 282\ \text{K}$

Answer:

$\boldsymbol{282\ \text{K}}$