Answered

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What is the temperature increase of 4.0 kg of water when it is heated by an 8.0 ^102 W immersion heater for exactly 10.0 min? (cp = 4186 J/kg·°C)

Sagot :

samuelpoly011

Answer:

ΔT = 28.667°C is the answer.

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

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