It took 400 calories of heat to raise 50 mL of water from 20 degrees Celsius to its final temperature. What was that temperature?
q = mass x specific heat x (Tfinal-Tinitial)
Solve for Tfinal.
Still trying to figure this out. I have read in my book and can't find anything. This question was actually on my first exam and I'm reviewing for my next one after the break. I don't understand how to set it up. 50 is my mass. Do I use the 400cal as my SH? I'm just lost on how to set it up.
To determine the final temperature of the water, we need to understand the concept of specific heat capacity. The specific heat capacity is the amount of heat energy required to raise the temperature of a substance by a certain amount.
In this case, we have 50 mL of water, which has a specific heat capacity of 1 calorie/gram °C. To convert the volume of water from milliliters (mL) to grams (g), we need to know the density of water. Since the density of water is approximately 1 gram/milliliter, we can assume that 50 mL of water is equivalent to 50 grams.
Given that it took 400 calories of heat energy to raise the temperature of 50 grams of water, we can apply the formula:
Q = m * c * ΔT
where:
Q is the heat energy (calories),
m is the mass of the substance (grams),
c is the specific heat capacity (calories/gram °C), and
ΔT is the change in temperature (final temperature - initial temperature).
In this case, we know that:
Q = 400 calories,
m = 50 grams, and
c = 1 calorie/gram °C.
Let's rearrange the formula to solve for ΔT:
ΔT = Q / (m * c)
Substituting the known values:
ΔT = 400 calories / (50 grams * 1 calorie/gram °C)
ΔT = 400 / 50 °C
ΔT = 8 °C
Therefore, the temperature of the water increased by 8°C. To find the final temperature, we need to add this change to the initial temperature of 20°C:
Final temperature = Initial temperature + ΔT
Final temperature = 20°C + 8°C
Final temperature = 28°C
Hence, the final temperature of the water is 28 degrees Celsius.