You go to kitchen to boil water. You pour 280 gram of water to a container, and put the container on a 293 W electric heater. Assume that all the heat from the heater is used to heat the water. The specific heat of water is 4.19 J/g ° C. As always, be accurate to 4 significant digits.

(a) If the temperature of the water is 21.6 ° C before you turn on the heater, how long does it take for the water temperature to reach 100 ° C after you turn on the heater? Provide your answers in minutes.
1 min

(b) After the water temperature reached 100 ° C, if you let the water keep boiling with the heater turn on for 20 minutes, how much water will be left in the container? Latent heat, or the heat of vaporization of water is 2260 J/g. Assume that all the vaporized water is removed from the container.

Last Question I will ask for the semester. Please help.

(a) You want to heat the water

delta T = 100-21.6 = 78.4 degrees.
The energy needed is
Q = C M *delta T
= 4.19*280*78.4 = 91,980 joules
Power*Time = 91,980
Time (in seconds) = 91,980/293 = 314 s

Convert that to minutes.

(b) After that length of time, additional heat added is used to vaporize. In 20 minutes, the added heat is 1200 seconds * 293 J/s = 351,600 J.

Divide that by the heat of vaporization to get the number of grams of water vaporized. Subtract that from 280 g to get the number of grams left.

Thank you Dr. DRWLS

To solve this problem, we need to use the formula for heat transfer:

Q = m * c * ΔT

Where:
Q = heat transfer (in joules)
m = mass of the water (in grams)
c = specific heat of water (in J/g °C)
ΔT = change in temperature (in °C)

Let's start with (a):

(a) To calculate the time required for the water temperature to reach 100 °C, we first need to find the amount of heat required to raise the temperature from 21.6 °C to 100 °C.

First, let's calculate the heat transfer using the formula mentioned above:

Q = m * c * ΔT

Q = 280 g * 4.19 J/g °C * (100 °C - 21.6 °C)

Q = 280 g * 4.19 J/g °C * 78.4 °C

Q ≈ 92844.48 J

Now, we know that the power of electric heater is 293 W. Power is defined as the rate of energy transfer. Therefore, we can use the following equation to find the time:

Q = P * t

Where:
Q = heat transfer (in joules)
P = power (in watts)
t = time (in seconds)

Let's convert the time from minutes to seconds:

1 minute = 60 seconds

Now, we can calculate the time required:

92844.48 J = 293 W * t

Solving for t:

t ≈ 316.929 seconds

To convert to minutes:

t ≈ 316.929 s * (1 min / 60 s)

t ≈ 5.282 minutes

So, it takes approximately 5.282 minutes for the water temperature to reach 100 °C after you turn on the heater.

Now, let's move on to (b):

(b) After the water temperature reached 100 °C, the remaining water will undergo a phase change from liquid to gas (vaporization). We need to calculate the mass of the water that has vaporized.

Using the formula:

Q = m * ΔHvap

Where:
Q = heat transfer (in joules)
m = mass of the water that has vaporized (in grams)
ΔHvap = heat of vaporization (in J/g)

Given that the heat transfer is the same as in part (a):

Q = 92844.48 J

We can rearrange the formula to solve for m:

m = Q / ΔHvap

m = 92844.48 J / 2260 J/g

m ≈ 41.07 g

So, approximately 41.07 g of water will have vaporized.

To find the amount of water remaining in the container, subtract the mass that has vaporized from the initial mass:

Remaining water = Initial mass - Mass vaporized

Remaining water = 280 g - 41.07 g

Remaining water ≈ 238.93 g

Therefore, after 20 minutes of boiling with the heater turned on, approximately 238.93 grams of water will be left in the container.