Answer:
35.24°C.
Step-by-step explanation:
To solve this problem, we can use the principle of conservation of energy. The heat lost by the water will be equal to the heat gained by the aluminum pan.
The equation for heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's calculate the heat lost by the water:
Q_water = m_water * c_water * ΔT_water
We know the mass of the water is 752 mL, which is equivalent to 752 grams (since the density of water is 1 g/mL).
m_water = 752 g
c_water = 4.18 J/g°C (specific heat of water)
ΔT_water = final temperature - initial temperature = 39°C - initial temperature
Similarly, let's calculate the heat gained by the aluminum pan:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
The mass of the aluminum pan is given as 1.00 kg.
m_aluminum = 1,000 g
c_aluminum = 0.9025 J/g°C (specific heat of aluminum)
ΔT_aluminum = final temperature - initial temperature = 39°C - 26°C
Since the heat lost by the water is equal to the heat gained by the aluminum pan, we can set up the equation:
Q_water = Q_aluminum
m_water * c_water * ΔT_water = m_aluminum * c_aluminum * ΔT_aluminum
Plugging in the values:
752 g * 4.18 J/g°C * (39°C - initial temperature) = 1,000 g * 0.9025 J/g°C * (39°C - 26°C)
Simplifying:
3135.776 g * (39°C - initial temperature) = 903.5 g * 13°C
122,287.664 g°C - 3135.776 g * initial temperature = 11,705.5 g°C
-3135.776 g * initial temperature = 11,705.5 g°C - 122,287.664 g°C
-3135.776 g * initial temperature = -110,582.164 g°C
Dividing by -3135.776 g:
initial temperature = (-110,582.164 g°C) / (-3135.776 g)
initial temperature ≈ 35.24°C
Therefore, the initial temperature of the water was approximately 35.24°C.
Hope this helps ^^