What is the temperature increase of 4.0 kg of water when heated by an 800-W immersion heater for 10 min? (c_w = 4 186 J/kg*$\degree$C) A)56 $\degree$C B)51 $\degree$C C)29 $\degree$C D)14 $\degree$C

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of energy transferred to the water:
Power = 800 W
Time = 10 min = 600 s
Energy = Power x Time = 800 x 600 = 480,000 J

Next, we can use the formula:
Energy = mass x specific heat capacity x temperature change

We know the mass is 4.0 kg and the specific heat capacity is 4,186 J/kg°C. We don't know the temperature change, but we can solve for it:

Temperature change = Energy / (mass x specific heat capacity) = 480,000 / (4.0 x 4,186) = 28.8°C

So the temperature of the water increases by 28.8°C. Converting to Fahrenheit gives 51.8°F, which is closest to answer C (29°C or 84.2°F).

What Is the Temperature Increase of 4 °\degree° C) A)56 °\degree° C B)51

What Is the Temperature Increase of 4 $\degree$ C)
A)56 $\degree$ C
B)51