2. In a calorimeter, 40.0g water at room temperature (20.4 C) was mixed with 60.0 g of water that was
initially 48.9 C. The resulting mixture reached a maximum temperature of 37.3 C. Calculate each of the
following.
a) The heat released by the hot water.
b) The heat absorbed by the room temperature water.
c) The calorimeter constant.
It doesn't help to post essentially the same question under another screen name.
For part A) I know it goes something like this. 40.0g(4.184)(48.9-37.3). But it asks for heat released by hot water i'm not exactly sure what to do.
I know c is A)-B)
See if you can do the others with the help of the previous problem.
To solve this problem, we need to use the formula for heat transfer:
q = m * c * ΔT
Where:
q is the heat transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C), and
ΔT is the change in temperature (in °C).
a) The heat released by the hot water:
To calculate this, we need to use the mass of the hot water and the specific heat capacity of water.
m_hot = 60.0 g (mass of hot water)
c_water = 4.18 J/g·°C (specific heat capacity of water)
ΔT_hot = (final temperature - initial temperature) = (37.3 °C - 48.9 °C) = -11.6 °C (the negative sign indicates a decrease in temperature)
Using the formula:
q_hot = m_hot * c_water * ΔT_hot
= 60.0 g * 4.18 J/g·°C * -11.6 °C
≈ -2906.08 J ≈ -2906 J
Therefore, the heat released by the hot water is approximately -2906 J.
b) The heat absorbed by the room temperature water:
To calculate this, we need to use the mass of the room temperature water and the specific heat capacity of water.
m_cold = 40.0 g (mass of room temperature water)
c_water = 4.18 J/g·°C (specific heat capacity of water)
ΔT_cold = (final temperature - initial temperature) = (37.3 °C - 20.4 °C) = 16.9 °C
Using the formula:
q_cold = m_cold * c_water * ΔT_cold
= 40.0 g * 4.18 J/g·°C * 16.9 °C
≈ 2831.872 J ≈ 2832 J
Therefore, the heat absorbed by the room temperature water is approximately 2832 J.
c) The calorimeter constant:
The calorimeter constant represents the amount of heat absorbed by the calorimeter when there is a 1°C change in temperature. It can be calculated by equating the heat absorbed by the room temperature water to the heat absorbed by the calorimeter.
q_cold = q_calorimeter
Using the formulas for each:
m_cold * c_water * ΔT_cold = q_calorimeter
We can substitute the known values:
40.0 g * 4.18 J/g·°C * 16.9 °C = q_calorimeter
q_calorimeter = 2832 J
Therefore, the calorimeter constant is 2832 J.
A)60.0g(4.814)(48.9-37.3)=3350.544
B)40.0(4.814)(20.4-37.3)=3254.264
C)A-B= 96.28