# Very difficult Chemistry Problem? Can someone help? ?

A bowl of 567 g of water is placed in a microwave oven that puts out 703 watts (J/s). How long would it take in seconds to increase the temperature of the water from 12.0oC to 54.4oC? Use SF.

Relevance
• 2 months ago

P = Q / t

t = Q / P = (m c deltaT) / P

t = (567 g * 4.186 J/g C° * 42.4 C °) / 703J/s4

t = 143 s or 2min 23s

When you get a good response,

This is the only reward we get.

• 1 month ago

Bit off topic but,

The time to heat up water (or any substance) in a microwave depends on its position and height inside the chamber. Because, the micro-waves are concentrated at specific heights and positions and this differs for every manufacturer.

Modern microwaves emit the waves from the top ceiling via a fan that scatters it in multiple directions; some manufacturers place the emitter and fan in the middle while others place it in a corner or side. Older microwaves had the emitter on the side wall, often without a fan; so if you piled food on a plate, the food on the top would get hotter faster than food at the bottom of the plate.

Tell this to your teacher/instructor, if you want to be a smart-***.

• 2 months ago

Q = mcDeltaT

Q = 567 g x 4.12J/(gK) x ( 54.4 - 12.0)

Q = 567 x 4.12 x 42.4 J

Q = 99,048 J  is the required heat

Hence time(s) = 99,048 / 703 = 140.89 s = 2mins 20 sec.

• Dr W
Lv 7
2 months ago

same troll different day.

• 2 months ago

I hope you never see an actual difficult problem.  ps, this is physics...

specific heat of water is 4.186 kJ/kgC = 4.186 J/gC

= 1 calorie/gC = 5.375 J/mol·K = 1 Btu/lb-F

energy required = 4.186 J/gC x 567 g x (54.4–12) = 100600 J

703 watts is 703 J/s

100600 J / 703 J/s = 143 sec