Question

1. a sample of water absorbs 3.00 × 10³ j of heat energy, and its temperature rises from 20.0°c to 31.2°c. what is the mass of the water? (1pts) heat = mass × specific heat × δt where δt = t_final - t_initial

□ 3.21 g
□ 64.0 g
□ 3.58 g
□ 1120 g

Explanation:

Step1: Calculate temperature change

$\Delta t = t_{final} - t_{initial} = 31.2^\circ\text{C} - 20.0^\circ\text{C} = 11.2^\circ\text{C}$

Step2: Rearrange heat formula for mass

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

Step3: Substitute known values

Specific heat of water is $4.184\ \frac{\text{J}}{\text{g}^\circ\text{C}}$, heat = $3.00 \times 10^3\ \text{J}$
$mass = \frac{3.00 \times 10^3\ \text{J}}{4.184\ \frac{\text{J}}{\text{g}^\circ\text{C}} \times 11.2^\circ\text{C}} \approx 64.0\ \text{g}$

Answer:

$\boldsymbol{64.0\ \text{g}}$ (Option: $\boldsymbol{\square\ 64.0\ \text{g}}$)