Question

150g of water cools from 50 degree celcius to 25 degree celcius in 5 minutes . calculate the rate of loss of heat .​

Answer 1

Answer:

Mass of water m = 50 g

Initial temperature of water (Ti) = 50°C

Final temperature of water (Tf) = 50°C

Specific heat of water = 4.2 J/g –°C

Time Δt = 5 minutes

= 5 × 60sec.

Heat lost by water ΔQ = msΔT

= ms (Tf – Ti)

= 50 × 4.2 × (5 – 50)

= – 50 × 4.2 × 45

= – 9450 J

Negative sign shows heat is lost by water.

Rate of heat lost