What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (cw = 4 186 J/kg⋅°C)

a. 56°Cb. 51°Cc. 29°Cd. 14°C

option C

Explanation:

given,

mass of water = 4 Kg

Water is heated to = 800 W

time of immersion = 10 min

= 10 x 60 = 600 s

using equation of specific heat

Q = m S ΔT

S is the specific heat capacity of water which is equal to 4182 J/kg°C.

and another formula of heat

Q = Pt

now,

P t = m S ΔT

800 x 600 = 4 x 4182 x ΔT

ΔT = 29° C

temperature increased is equal to ΔT = 29° C

Hence, the correct answer is option C