Question

in your secret underground labatory, you created water vapor out of liquid water by adding 615 calories of heat. if you had 1 gram of sample to start with, what was the strarting temperature of the liquid? write your answer as ## degrees C.​

Answer 1

From the statements in the question, the starting temperature before 615 calories of heat was added is 24 °C.

Given that the total energy added to the water = Heat added to raise the temperature of the water to boiling point + latent heat of vaporization of water

Mass of the water = 1 g

Boiling point of water = 100°C

Starting temperature of the water = θ°C

Latent heat of vaporization of water = 539 cal/g

Specific heat capacity of water = 1 cal/g°C

Hence;

H= mc(100 -θ) + mL

H = m(c(100 -θ) + L)

H = 1[1 × (100 - θ) + 539)]

615 = 100 - θ +  539

θ = 100 + 539 - 615

θ = 24 °C

Learn more: brainly.com/question/1453843