75.0 claories of heat were added to 50.0 ml of water, the initial temperature of the water being 25.0 celcius. What is the final temp. of water?
To determine the final temperature of the water, we can use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy added to the water (in calories),
m is the mass of water (in grams),
c is the specific heat capacity of water (approximately 1 calorie/gram°C), and
ΔT is the change in temperature (final temperature - initial temperature) of the water (in °C).
Given:
Q = 75.0 calories
m = 50.0 ml of water (which can be converted to grams since the density of water is 1 g/ml)
c = 1 calorie/gram°C
Initial temperature (T1) = 25.0°C
First, convert the volume of water to mass using the density of water. Since the density of water is 1 g/ml, 50.0 ml of water is equal to 50.0 grams.
Next, rearrange the equation to solve for ΔT:
ΔT = Q / (m * c)
Substitute the given values:
ΔT = 75.0 calories / (50.0 grams * 1 calorie/gram°C)
Simplify:
ΔT = 75.0 / 50.0 °C
ΔT = 1.5°C
Now, find the final temperature (T2):
T2 = T1 + ΔT
Substitute the values:
T2 = 25.0°C + 1.5°C
T2 = 26.5°C
Therefore, the final temperature of the water is 26.5°C.