Use the heating curve of 50 g of water and the list of values to answer the question.
specific heat of ice = 2.10 J/(g·°C)
specific heat of water = 4.18 J/(g·°C)
specific heat of water vapor = 2.07 J/(g·°C)
latent heat of fusion of ice = 333.4 J/g
latent heat of vaporization of water = 2256 J/g
If 9200 J of heat were added to the water as depicted by the heating curve, what is the final temperature of water?
its 54 C
I don't know what the heating curve looks like and I don't know what that has to do with the problem.
I tried 34 C and I got it wrong...
Multiply the specific heat of water by the grams of water you are given.
(4.18J/g*C)(50g)=209J/C
Then, divide 9200J by 209J/C to get 44C
it wasn't 44 either
To determine the final temperature of the water, we need to break down the heating process into different stages. Let's go step by step:
Stage 1: Heating ice from -10°C to 0°C (heating solid ice)
First, we need to determine the amount of heat required to raise the temperature of the ice from -10°C to 0°C. The specific heat capacity of ice is 2.10 J/(g·°C), and the mass of ice is 50 g.
ΔQ = m * c * ΔT
ΔQ = 50 g * 2.10 J/(g·°C) * (0°C - (-10°C))
ΔQ = 50 g * 2.10 J/(g·°C) * 10°C
So, the heat required to raise the temperature of ice from -10°C to 0°C is 1050 J.
Stage 2: Melting ice at 0°C (latent heat of fusion)
Once the ice reaches 0°C, we need to calculate the amount of heat required to melt the ice without changing its temperature. The latent heat of fusion of ice is 333.4 J/g.
ΔQ = m * L
ΔQ = 50 g * 333.4 J/g
So, the heat required to melt the ice is 16,670 J.
Stage 3: Heating water from 0°C to final temperature (heating liquid water)
After the ice has melted, we are left with liquid water that needs to be heated from 0°C to the final temperature. The specific heat capacity of water is 4.18 J/(g·°C).
ΔQ = m * c * ΔT
ΔQ = 50 g * 4.18 J/(g·°C) * (final temperature - 0°C)
Stage 4: No phase changes occur at the final temperature or after, so we don't need to consider latent heats anymore.
Adding up the heat from each stage:
Total heat = 1050 J + 16,670 J + 50 g * 4.18 J/(g·°C) * (final temperature - 0°C)
Since we know that a total of 9200 J of heat were added, we can set up an equation:
9200 J = 1050 J + 16,670 J + 50 g * 4.18 J/(g·°C) * (final temperature - 0°C)
Simplifying the equation:
9200 J = 17,720 J + 50 g * 4.18 J/(g·°C) * final temperature
Rearranging the equation to solve for the final temperature:
50 g * 4.18 J/(g·°C) * final temperature = 9200 J - 17,720 J
50 g * 4.18 J/(g·°C) * final temperature = -7820 J
final temperature = (-7820 J) / (50 g * 4.18 J/(g·°C))
final temperature ≈ -9.42°C
Based on the given values and calculations, the final temperature of the water is approximately -9.42°C.