A 1.50 X 10^2g piece of brass(specific heat capacity 3.80 X 10^2 J(kgC) is submerged in 400 mL of water at 27.7 Degrees Celsius. What is the original temperature of the brass if the mixture has a temperature of 28.0 degrees Celsius
haat added to brass+heat added to water=0
150*cbrass*(28-Ti)+400*cwater*(28-27.7)=0
solve for Ti
Michelle/Rosie -- please use the same name for your posts.
absolutely several thousand off.
To determine the original temperature of the brass, we can use the principle of heat transfer, which states that the heat gained by the water is equal to the heat lost by the brass.
The equation for the heat transfer is given by:
Q_water = -Q_brass
Where:
Q_water is the heat gained by water
Q_brass is the heat lost by brass
The heat gained or lost in a substance can be calculated using the formula:
Q = m * c * ∆T
Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat capacity of the substance
∆T is the change in temperature
Let's calculate the heat gained by the water:
Q_water = m_water * c_water * ∆T_water
Where:
m_water = mass of water = volume of water * density of water
c_water = specific heat capacity of water = 4.18 J/(g°C)
∆T_water = change in temperature of water = final temperature - initial temperature
First, convert the volume of water from milliliters (mL) to grams (g):
mass_water = volume_water * density_water
The density of water is approximately 1 g/mL.
mass_water = 400 mL * 1 g/mL
mass_water = 400 g
∆T_water = 28.0°C - 27.7°C
∆T_water = 0.3°C
Now we can calculate the heat gained by the water:
Q_water = 400 g * 4.18 J/(g°C) * 0.3°C
Q_water = 500.4 J
Since the heat gained by the water is equal to the heat lost by the brass, we can set up the equation:
Q_water = -Q_brass
500.4 J = -m_brass * c_brass * ∆T_brass
To solve for ∆T_brass, we need to calculate the mass of the brass (m_brass):
m_brass = 1.50 X 10^2 g
Now we can rearrange the equation to solve for ∆T_brass:
∆T_brass = -500.4 J / (m_brass * c_brass)
∆T_brass = -500.4 J / (1.50 X 10^2 g * 3.80 X 10^2 J/(kg°C))
∆T_brass = -0.877°C
Finally, we can now find the original temperature of the brass:
Original temperature of brass = Temperature of mixture + ∆T_brass
Original temperature of brass = 28.0°C + (-0.877°C)
Original temperature of brass = 27.123°C
Therefore, the original temperature of the brass is 27.123°C.