Question

How much heat is required to convert 95.0 grams of water at 30.0 °C to 75.0 °C under 1 atm?

Answer 1

17847 joules

Step-by-step explanation:

m = 95.0 grams (mass of water)

c = 4.18 J/g°C (specific heat capacity of water)

ΔT = 75.0°C - 30.0°C = 45.0°C

Q = m * c * ΔT

Q = 95.0 g * 4.18 J/g°C * 45.0°C

Q = 17847 J