Question

how much energy is absorbed when 100. g of water goes from 0.0°c to 100.0°c? the specific heat of water is 4.184 j/g°c.
q = ? j

Explanation:

Step1: Recall the formula for heat energy

The formula to calculate the heat energy \( q \) absorbed or released is \( q = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Identify the values

Given \( m = 100. \, \text{g} \), \( c = 4.184 \, \text{J/g}^\circ\text{C} \), initial temperature \( T_i = 0.0^\circ\text{C} \), final temperature \( T_f = 100.0^\circ\text{C} \). Calculate \( \Delta T = T_f - T_i = 100.0^\circ\text{C} - 0.0^\circ\text{C} = 100.0^\circ\text{C} \).

Step3: Substitute the values into the formula

Substitute \( m = 100. \, \text{g} \), \( c = 4.184 \, \text{J/g}^\circ\text{C} \), and \( \Delta T = 100.0^\circ\text{C} \) into \( q = mc\Delta T \).
\( q = 100. \, \text{g} \times 4.184 \, \text{J/g}^\circ\text{C} \times 100.0^\circ\text{C} \)

Step4: Calculate the result

First, multiply \( 100. \) and \( 4.184 \): \( 100. \times 4.184 = 418.4 \). Then multiply by \( 100.0 \): \( 418.4 \times 100.0 = 41840 \).

Answer:

\( 41840 \)