Answer:
23,445 joules of heat are required to boil 75 grams of water, considering it is initially at room temperature (25°C).
Step-by-step explanation:
To calculate the amount of heat required to boil 75 grams of water, we need to know the specific heat capacity of water and the temperature change involved. For water, the specific heat capacity is 4.18 J/(g·°C). To boil water, we need to raise its temperature from room temperature to its boiling point (100°C). Assuming room temperature to be 25°C, the temperature change is 100°C - 25°C = 75°C.
Using the formula for calculating heat, where q is the heat required, m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change:
q = m × c × ΔT
Substituting the values:
q = 75 g × 4.18 J/(g·°C) × 75°C
q = 75 × 4.18 × 75
q = 23445 J
So, 23,445 joules of heat are required to boil 75 grams of water, considering it is initially at room temperature (25°C). Note that this calculation assumes no heat loss to the environment and does not account for the heat required to change water from liquid to vapor (latent heat of vaporization).