Certainly! Let's tackle this problem step-by-step.
1. Identify given values:
- Mass of water, [tex]\(m\)[/tex] = 5 kg
- Initial temperature of water, [tex]\( T_{\text{initial}} \)[/tex] = [tex]\(20^{\circ} C\)[/tex]
- Final temperature of water, [tex]\( T_{\text{final}} \)[/tex] = [tex]\(100^{\circ} C\)[/tex]
- Specific heat capacity of water, [tex]\( c \)[/tex] = 1000 cal/kg°C
2. Determine the temperature change:
[tex]\[
\Delta T = T_{\text{final}} - T_{\text{initial}}
\][/tex]
Substituting the given values:
[tex]\[
\Delta T = 100^{\circ} C - 20^{\circ} C = 80^{\circ} C
\][/tex]
3. Apply the formula for heat energy:
The formula to calculate the heat required ([tex]\(Q\)[/tex]) is:
[tex]\[
Q = m \cdot c \cdot \Delta T
\][/tex]
Substituting the given values and the calculated temperature change:
[tex]\[
Q = 5 \text{ kg} \times 1000 \text{ cal/kg°C} \times 80^{\circ} C
\][/tex]
Simplifying this, we get:
[tex]\[
Q = 5 \times 1000 \times 80 = 400000 \text{ cal}
\][/tex]
So, the quantity of heat required to raise the temperature of 5 kg of water from [tex]\(20^{\circ} C\)[/tex] to [tex]\(100^{\circ} C\)[/tex] is 400000 calories.
