Given that the specific heat of water is 4.18 kJ / (kg·°C), how much heat does it take to raise the temperature of 3.5 kg of water from 25°C to 55°C?

The heat required = 438900 J

Explanation:

mass = 3.5 Kg

change in temperature=ΔT=55-25=30°C

specific heat of iron = 4.18 kJ / (kg·°C) = 4180 J / (kg·°C)

the formula for heat is given by

Q = m C ΔT

Q=3.5 (4180) (30)

Q=438900 J