Physics 12.32 help please 4

Question

Suppose the 620W of radiation emitted in a microwave oven is absorbed by a very lightweight cup holding 260g of water.

Approximately how long will it take to heat the water from 20 degrees C} to 70degrees C?

Answer 1

This is a fun problem to pay attention to units.

Start with the equation

Q=mcT

Q=energy change= ? in Joules

m=mass = 260 g (assume the cup itself is massless and retains no heat)

c=specific heat capacity= 4.184J/(g*K) for water

T=temperature change= 70C-20C= 50C= 50K

Q=260g*4.184J/(g*K)*50K= 54392 Joules

which is how much energy is needed to raise the temperature of the water in the cup

Now, lets see how long it'll take. We know the rate of energy input so this should be easy. We can just look at the units to see what to do:

1)We need 54392 joules of energy. 1 J= 1 N*m = 1 Watt*sec

2) We have the rate= 620 W= 630 J/sec

3) We note that since Watts=Joules/Second then

                                    Seconds*Watts=Joules  and

                                   Seconds= Joules/watts

4) So time=total energy/rate of input . Makes sense, right? And it agrees with our units, so we know we're right. Also:

J/W= J/(J/s) = s

5) time=54392 Joules/620Watts = 87.73 seconds

Wasn't that fun?

Answer 2

Total energy from microwave = Energy absorbed by water

620T = MCp(delta T)

where

T = time for water to be heated from 20 deg to 70 deg

M = mass of water = 260 grams (given)

Cp =specific heat of water = 4.184 joules/g C

delta T = 70 - 20 = 50

Substituting appropriate values,

620T = 260(4.184)(50)

T = 260(4.184)(50)/620

T = 87.73 seconds

T = 1 minute 28 seconds (approximately)