You have a bottle of water. The water has a mass of 29 grams. You heat the water with 102 calories of heat and the final temperature of the water is 43°C. What was the initial temperature (in Celsius) of the water before you added the heat? The specific heat of water is 1 calorie / (°Cg).
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
You know q, mass H2O, specific heat and Tfinal. Solve for Tinitial.
To find the initial temperature of the water before adding heat, we can use the equation:
Q = mcΔT
where Q is the heat energy absorbed or released by the water, m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.
In this case, we have:
Q = 102 calories
m = 29 grams
c = 1 calorie / (°Cg)
ΔT = final temperature - initial temperature = 43°C - initial temperature
Let's plug in the values into the equation:
102 calories = 29 grams * 1 calorie / (°Cg) * (43°C - initial temperature)
Next step is to solve for the initial temperature:
102 calories = 29 grams * 1 calorie / (°Cg) * (43°C - initial temperature)
Divide both sides by (29 grams * 1 calorie / (°Cg)):
102 calories / (29 grams * 1 calorie / (°Cg)) = 43°C - initial temperature
Simplify the units on the left side:
102 g°C = 43°C - initial temperature
Rearrange the equation:
initial temperature = 43°C - 102 g°C
Compute the final answer:
initial temperature = 43°C - 102 g°C = -59°C
Therefore, the initial temperature of the water was -59°C before adding the heat.