You and your friend went to kitchen to boil water. The specific heat of water is

Question

You and your friend went to kitchen to boil water. The specific heat of water is 4.19 J/g °C, and the latent heat of vaporization of water is 2260 J/g.

(a) You poured 211 gram of water to a container and put the container on a 279 W electric heater. After turning on the heater, you went away for 20 minutes to do something else. The container was not covered so that the vaporized water was removed from the container. When you came back to the kitchen 20 minutes later to check the water, the container had 93.2 gram of water left in it boiling. What was the temperature of your water when when you turned on the heater?
°C

(b) Your friend wanted her own boiling water. She poured 344 gram of water into another container and put her own electric heater at the same time as you did. The initial temperature of her water was the same as yours. When you came back to the kitchen to check the water in 20 minutes, there was 236.8 gram of water left in her container. What was the power of your friend's electric heater?
W

Explanation / Answer

Here ,

a)

initial mass of water , m = 211 gm

mass of water left , m1 = 93.2 gm

let the iniital temperature of water was Ti degree C

heat transferred to water = Power * time

m * 4.19 * (100 - Ti) + (m - m1) * 2260 = Power * time

211 * 4.19 * (100 - Ti) + (211 - 93.2) * 2260 = 279 * 20 * 60

solving for Ti

Ti = 22.44 degree C

the initial temperature of water was 22.444 degree C

b)

Here, let the power of heater is P

heat transferred to water = Power * time

344 *4.19 * (100 - 22.44) + (344 - 236.8) * 2260 = P * 20 * 60

solving for P

P = 295 W

the power of the heater is 295 W

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.