Water (2330 g) is heated until it just begins to boil. If the water absorbs 5.53×105 J of heat in the process, what was the initial temperature of the water?
Q=mc∆t
Q = heat energy
m = mass
c = specific heat (of water)
∆t= change in temperature (= final - initial temp)
final temp= boiling point of water i.e. 100*C
c= 4.186 joule/gram °C
This is how far i got
2330g *4.184J/g degree C *(100 degree C-unk) =537,000J
so
9748.72J * unk =537,000J
so
537000/9748.72=55.08
so
100-55.08= Initial temp of 44.92 degrees celsius
:)
To find the initial temperature of the water, you can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed by the water
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
Q = 5.53 × 10^5 J
m = 2330 g
c = 4.18 J/g°C (specific heat capacity of water)
First, convert the mass of the water to kilograms:
m = 2330 g = 2.33 kg
Rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
ΔT = 5.53 × 10^5 J / (2.33 kg * 4.18 J/g°C)
Now, perform the calculation:
ΔT = 5.53 × 10^5 J / (2.33 kg * 4.18 J/g°C) ≈ 58.75 °C
Finally, subtract the change in temperature from the boiling point of water (100 °C) to find the initial temperature:
Initial temperature = 100 °C - 58.75 °C = 41.25 °C
Therefore, the initial temperature of the water was approximately 41.25 °C.
To determine the initial temperature of the water, we can use the equation:
Q = mcΔT
where:
Q is the amount of heat absorbed by the water,
m is the mass of the water,
c is the specific heat capacity of water, and
ΔT is the change in temperature.
We know:
Q = 5.53×10^5 J (the amount of heat absorbed)
m = 2330 g (the mass of the water)
c = 4.18 J/g·°C (specific heat capacity of water)
We have to rearrange the equation to solve for ΔT:
ΔT = Q / (mc)
Substituting the known values:
ΔT = (5.53×10^5 J) / (2330 g * 4.18 J/g·°C)
Now we can calculate ΔT:
ΔT ≈ 56.49 °C
The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. Since the water just begins to boil, its final temperature is 100 °C at sea level. Therefore, we can calculate the initial temperature (T_i) as:
T_i = T_final - ΔT
T_i = 100 °C - 56.49 °C ≈ 43.51 °C
So, the initial temperature of the water was approximately 43.51 °C.