If the temperature of 50.0 grams of water was raised to 50°c by the addition of 4180J of heat energy. What was the initial temperature of the water?
q = 4180 = mass H2O x specific heat H2O x (Tfinal0Tiitial)
specific heat H2O is 4.18 J/oC
Q = m • C • ΔT
Q = 4180J
m = 50.0g
C = 4.184 J/g*C
ΔT = (50*C - T(initial))
4180 = 50.0 • 4.184 • (50 - T)
4180 = 209.2 (50 - T)
4180 = -209.2T + 10460
-209.2T = -6280
T(initial) = 30.0*C
To find the initial temperature of the water, we can use the principle of heat transfer:
Q = mc∆T
Where:
Q = heat energy in Joules (4180 J)
m = mass of water in grams (50.0 g)
c = specific heat capacity of water (4.18 J/g°C)
∆T = change in temperature (final temperature - initial temperature)
Rearranging the equation, we get:
∆T = Q / (mc)
Substituting the given values, we have:
∆T = 4180 J / (50.0 g * 4.18 J/g°C)
∆T = 19.9524°C
Since the final temperature is 50°C, we can find the initial temperature by subtracting the change in temperature from the final temperature:
Initial temperature = Final temperature - ∆T
Initial temperature = 50°C - 19.9524°C
Initial temperature ≈ 30.05°C
Therefore, the initial temperature of the water was approximately 30.05°C.
To find the initial temperature of the water, we can use the formula:
Q = mcΔT
Where:
Q = heat energy added (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
In this case, we know:
Q = 4180 J
m = 50.0 g
c = 4.18 J/g°C
ΔT = 50°C (final temperature - initial temperature)
Plugging in these values, we get:
4180 J = (50.0 g) * (4.18 J/g°C) * (50°C - initial temperature)
Simplifying the equation, we have:
4180 J = 209 J/°C * (50°C - initial temperature)
Divide both sides of the equation by 209 J/°C:
4180 J / 209 J/°C = 50°C - initial temperature
Simplifying further, we have:
20 = 50 - initial temperature
Now, subtract 50 from both sides:
20 - 50 = - initial temperature
Simplifying, we get:
-30 = - initial temperature
Finally, multiply both sides by -1 to solve for the initial temperature:
initial temperature = 30°C
Therefore, the initial temperature of the water was 30°C.