A sample of 35.5 grams of liquid water is cooled from boiling point ( 100 °C). Approximately 3713 Joules of heat are released in this process. What is the final temperature of the liquid water? Liquid water has a specific heat of 4.184 J/g .
Question
A sample of 35.5 grams of liquid water is cooled from boiling point ( 100 °C). Approximately 3713 Joules of heat are released in this process. What is the final temperature of the liquid water? Liquid water has a specific heat of 4.184 J/g .
Solution
To solve this problem, we can use the formula for heat transfer:
q = mcΔT
where: q = heat energy (in Joules) m = mass (in grams) c = specific heat (in Joules per gram per degree Celsius) ΔT = change in temperature (in degrees Celsius)
We know that q = -3713 J (the negative sign indicates heat loss), m = 35.5 g, and c = 4.184 J/g°C. We're trying to find ΔT, which we can solve for by rearranging the formula:
ΔT = q / (mc)
Substituting the known values gives us:
ΔT = -3713 J / (35.5 g * 4.184 J/g°C) = -25°C
This is the change in temperature. Since the water was initially at 100°C, the final temperature is:
100°C - 25°C = 75°C
So, the final temperature of the liquid water is approximately 75°C.
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