Questions: How much energy (in kJ) is required to raise the temperature of 219 g of water by 75.0°C?

Solution Steps

We are given:

• Mass of water, \( m = 219 \, \text{g} \)
• Temperature change, \( \Delta T = 75.0 \, ^\circ \text{C} \)
• Specific heat capacity of water, \( c = 4.184 \, \text{J/g} \cdot ^\circ \text{C} \)

Since the specific heat capacity is given in \(\text{J/g} \cdot ^\circ \text{C}\), we can keep the mass in grams: \[ m = 219 \, \text{g} \]

The formula to calculate the energy required to raise the temperature is: \[ Q = mc\Delta T \] Substituting the given values: \[ Q = 219 \, \text{g} \times 4.184 \, \text{J/g} \cdot ^\circ \text{C} \times 75.0 \, ^\circ \text{C} \]

\[ Q = 219 \times 4.184 \times 75.0 \] \[ Q = 68685.3 \, \text{J} \]

Since \(1 \, \text{kJ} = 1000 \, \text{J}\): \[ Q = \frac{68685.3 \, \text{J}}{1000} \] \[ Q = 68.6853 \, \text{kJ} \]

Final Answer