A total of 506 J of heat are added to 50.0 g of water initially at 15.0 °C. What is the final temperature of the water? The specific heat of liquid water is 4.184 J/g °C.
50.0 * 4.184 * (t - 15.0) = 506
solve for t
To determine the final temperature of the water, we can use the equation:
\[q = m \cdot c \cdot \Delta T\]
Where:
- \(q\) is the heat added to the water (506 J in this case)
- \(m\) is the mass of the water (50.0 g in this case)
- \(c\) is the specific heat capacity of water (4.184 J/g °C in this case)
- \(\Delta T\) is the change in temperature we want to find
Rearranging the equation, we have:
\[\Delta T = \frac{q}{m \cdot c}\]
Plugging in the given values, we can calculate the change in temperature:
\[\Delta T = \frac{506 \, \text{J}}{50.0 \, \text{g} \cdot 4.184 \, \text{J/g} °C}\]
To find the final temperature of the water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy added to the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Rearranging the equation, we can solve for ΔT:
ΔT = Q / (m * c)
Now, let's plug in the values from the problem:
Q = 506 J
m = 50.0 g
c = 4.184 J/g °C
Substituting these values into the equation, we get:
ΔT = 506 J / (50.0 g * 4.184 J/g °C)
Now, we can solve for ΔT:
ΔT = 506 J / (209.2 g °C)
ΔT = 2.42 °C
Since the water was initially at 15.0 °C, the final temperature will be:
Final temperature = Initial temperature + ΔT
Final temperature = 15.0 °C + 2.42 °C
Final temperature = 17.42 °C
Therefore, the final temperature of the water is 17.42 °C.