Answer:
So, Manny would require [tex]\(83680 \, J\)[/tex] heat energy to raise the temperature of [tex]\(250 \, \text{ml}\)[/tex] of water from [tex]\(20°C\) to \(100°C\).[/tex]
Explanation:
To calculate the amount of heat required to raise the temperature of water from \(20°C\) to \(100°C\), we can use the formula:
[tex]\[ Q = mc\Delta T \][/tex]
Where:
[tex]- \( Q \) is the amount of heat energy (in joules, \( J \))[/tex]
[tex]- \( m \) is the mass of the substance (in grams, \( g \))[/tex]
[tex]- \( c \) is the specific heat capacity of the substance (in \( J/g°C \))[/tex]
[tex]- \( \Delta T \) is the change in temperature (in degrees Celsius, \( °C \))[/tex]
Given:
[tex]- \( m = 250 \, \text{ml} \) of water[/tex]
[tex]- \( c = 4.184 \, \text{J/g} \cdot °C \)[/tex]
[tex]- \( \Delta T = 100°C - 20°C = 80°C \)[/tex]
First, let's convert the volume of water from milliliters to grams. Since the density of water is [tex]\(1 \, \text{g/ml}\), \(250 \, \text{ml}\)[/tex] of water is equal to [tex]\(250 \, \text{g}\).[/tex]
Now, we can calculate the amount of heat:
[tex]\[ Q = (250 \, \text{g}) \times (4.184 \, \text{J/g} \cdot °C) \times (80°C) \][/tex]
[tex]\[ Q = 250 \times 4.184 \times 80 \, J \][/tex]
[tex]\[ Q = 83680 \, J \][/tex]
So, Manny would require [tex]\(83680 \, J\)[/tex] heat energy to raise the temperature of [tex]\(250 \, \text{ml}\)[/tex] of water from [tex]\(20°C\) to \(100°C\).[/tex]
