How much thermal energy is required to reduce 150 mL of tap water measured at 70C to room temperature at 21C?

To calculate the thermal energy required to cool 150 mL of water from 70°C to 21°C, we can use the formula ( Q = mc\Delta T ), where ( Q ) is the thermal energy, ( m ) is the mass of the water (150 g for 150 mL), ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature. The change in temperature is ( 70°C - 21°C = 49°C ). Thus, ( Q = 150 , \text{g} \times 4.18 , \text{J/g°C} \times 49°C \approx 30,735 , \text{J} ).