Prior to your weekend hike at Golden Gate Canyon State Park you filed your insul

Question

Prior to your weekend hike at Golden Gate Canyon State Park you filed your insulated water bottle with 351.2 g of water (at 25.0 degree C), which you cooled down by adding 63.8 g of ice cubes (at -18.0 degree C). After hiking for several hours you opened the insulated water bottle. Which of the following did you observe? (Assume no loss of heat to the surroundings.) 1. The ice melt. 2. There was some ice floating in the water. Your Task: Support your answer by determining the final temperature of the water then, explain your process in writing. Imagine that you are writing to a classmate who doesnt yet understand specific heat capacity and heat of fusion. Your task is to explain your reasoning so clearly that your microtheme serves as a little textbook, teaching your classmate the physical principles involved Thus, your microtheme will be judged not simply on whether or not your figure out the correct answer, but also on whether or not you can write clearly enough to teach a fellow classmate. If you think a diagram of a heating curve would help you can include it with your explanation.

Explanation / Answer

m = 251.2 g of water, T = 25°C

m = 63.8 g of ice T = -18°C

First, identify total content of Ice energy

For ice: sensible heat of ice from -18 to 0 + latent heat of melting

Qsensible = m*C*(Tf-Ti) = 63.8 *2.01*(0--18) = 2308.284 J

Qlatent = m*LH = 63.8*334 = 21309.2 J

Qtotal = 2308.284 +21309.2 = 23617.484 J

For water:

Qsensible = m*C*(Tf-Ti) = 251.2*4.184*(25-0) = 26275.52 J

Qfreezing = m*LH = 251.2*2264.76= 568907.7 J

Therefore, there is much more heat in the water, expect all ice to MELT

proof:

26275.52 J of water will cool 23617.484 J

so expect this to happen:

Ice will go from -18°C to 0°C, then will melt at 0°C, then will convert to water and rise from 0°C to the equilbirium temperature Teq

the water will:

lower its temperature from 25°C to Teq, which is above of 0°C

the equation:

Qice = -Qwater

mice*Cice*(0-Ti) + m*melting energy + m*C*(Teq-0°C) = -mwater*Cwater*(Teq - 25°C)

substitute data

63.8 *2.01*(0--18) + 63.8 *334 +63.8 *4.184*(Teq-0°C) = -251.2*4.184*(Teq - 25)

Solve for Teq

23617.484+266.9392*Teq = -1051.0208*Teq + 26275.52

Teq*(266.9392+1051.0208) = 26275.52-23617.484

Teq = (26275.52-23617.484) / (266.9392+1051.0208)

Teq = 2.01 °C will be the final tempearture

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