Question

1. 100.0 g of 4.0°c water is heated until its temperature is 37°c. calculate the amount of heat energy needed to cause this rise in temperature.

Explanation:

Step1: Recall the heat formula

The formula for heat energy \( Q \) is \( Q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is temperature change. For water, \( c = 4.18 \, \text{J/g}^\circ\text{C} \).

Step2: Calculate temperature change

\( \Delta T = T_2 - T_1 = 37^\circ\text{C} - 4.0^\circ\text{C} = 33^\circ\text{C} \)

Step3: Substitute values into formula

\( m = 100.0 \, \text{g} \), \( c = 4.18 \, \text{J/g}^\circ\text{C} \), \( \Delta T = 33^\circ\text{C} \)
\( Q = 100.0 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 33^\circ\text{C} \)
\( Q = 100.0 \times 4.18 \times 33 \, \text{J} \)
\( Q = 13794 \, \text{J} \approx 1.4 \times 10^4 \, \text{J} \) (or 13.8 kJ)

Answer:

The heat energy needed is approximately \( 1.4 \times 10^4 \, \text{J} \) (or 13794 J, or 13.8 kJ).