Question

What is the temperature increase of 4.0 kg of water when it is heated by

Q= 4.80 * 10^5 J? (cp= 4186 J/kg C

Answer 1

ΔT = 28.667°C is the answer.

Step-by-step explanation:

Given, Mass of water = 4.0 kg

Water is heated by = 8 × 10² W = 800 W

Time = 10 min = 10 × 60 = 600 s

Now, using the equation of specific heat,

Q = mcΔT

where c is the specific heat capacity of water = 4186 J/kg°C

Q = Pt

Substituting,

Pt = mcΔT

800 × 600 = 4 × 4186 × ΔT

480000 = 16744 × ΔT

ΔT = 28.667

Temperature increased, ΔT = 28.667°C