What is the temperature increase of 3.00 kg of water when heated by a 900-W immersion heater for 9.00 min? (c_w = 4 186 J/kg⋅°C) A)77.4°C B)48.4°C C)38.7°C D)19.4°C E)0.645°C

Question

Quizplus · Accepted Answer

The amount of heat transferred to the water can be calculated using the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of water, c is the specific heat capacity, and ΔT is the temperature increase.

We have:

m = 3.00 kg
c = 4 186 J/kg⋅°C
ΔT = (900 W) × (9.00 min) × (60 s/min) / (3.00 kg × 4 186 J/kg⋅°C) = 38.7°C

Therefore, the temperature increase of the water is 38.7°C, so the answer is choice C.

What Is the Temperature Increase of 3