If 200.0 grams of water is to be heated from 24.0 C to 100.0 C to make a cup of tea how much heat must be added?

To calculate the heat required to heat 200.0 grams of water from 24.0 °C to 100.0 °C, we can use the formula ( q = mc\Delta T ), where ( m ) is the mass of the water (200.0 g), ( c ) is the specific heat capacity of water (approximately 4.18 J/g°C), and ( \Delta T ) is the change in temperature (100.0 °C - 24.0 °C = 76.0 °C). Therefore, ( q = 200.0 , \text{g} \times 4.18 , \text{J/g°C} \times 76.0 , \text{°C} ), which equals approximately 63440 J (or 63.4 kJ) of heat that must be added.