Question

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?

Answer 1

The heat required =438900 J

Step-by-step 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