95.0mLmL of H2O\rm H_2O is initially at room temperature (22.0∘C\rm{} ^{\circ}C). A chilled steel rod at 2.0∘C\rm ^{\circ}C is placed in the water. If the final temperature of the system is 21.3∘C^{\circ}C, what is the mass of the steel bar?
Specific heat of water = 4.18 J/g⋅∘C\rm J/g{\cdot}^{\circ}C
Specific heat of steel = 0.452 J/g⋅∘C
heat gained by rod + heat lost by H2O = 0
[mass rod x specific heat rod x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for mass rod
31.86
To solve this problem, we can use the equation Q = mcΔT, where Q is the heat transferred, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred by the water (Q_water):
Q_water = (mass_water) * (specific heat_water) * (ΔT_water)
We know that the volume of water is 95.0 mL, which is equivalent to 95.0 grams since the density of water is 1 g/mL.
Next, let's calculate ΔT_water:
ΔT_water = (final temperature_water) - (initial temperature_water)
ΔT_water = 21.3∘C - 22.0∘C
ΔT_water = -0.7∘C
Now, let's substitute the known values into the equation for Q_water:
Q_water = (95.0g) * (4.18 J/g⋅∘C) * (-0.7∘C)
Since the steel rod is chilled, it absorbs the heat transferred by the water. The steel rod gains the same amount of heat as the water loses. Therefore, Q_steel = -Q_water.
Next, let's calculate ΔT_steel:
ΔT_steel = (final temperature_steel) - (initial temperature_steel)
ΔT_steel = 21.3∘C - 2.0∘C
ΔT_steel = 19.3∘C
Now, let's calculate the mass of the steel rod (mass_steel) using the equation Q_steel = (mass_steel) * (specific heat_steel) * (ΔT_steel):
(mass_steel) = (Q_steel) / ((specific heat_steel) * (ΔT_steel))
Since Q_steel = -Q_water, we can substitute the known values into the equation and solve for (mass_steel):
(mass_steel) = -((95.0g) * (4.18 J/g⋅∘C) * (-0.7∘C))/ ((0.452 J/g⋅∘C) * (19.3∘C))
Calculating the mass_steel will give you the mass of the steel rod in grams.
To solve this problem, we can use the principle of heat transfer. The heat gained by the water is equal to the heat lost by the steel rod. We can use the formula:
Q = m * C * ΔT
Where:
Q is the heat gained or lost (in joules)
m is the mass of the substance (in grams)
C is the specific heat capacity of the substance (in J/g⋅∘C)
ΔT is the change in temperature (in ∘C)
Let's calculate the heat gained by the water:
Q_water = m_water * C_water * ΔT_water
Since the water is initially at 22.0∘C and the final temperature is 21.3∘C, we have:
ΔT_water = 21.3∘C - 22.0∘C = -0.7∘C
We know that the specific heat capacity of water (C_water) is 4.18 J/g⋅∘C, and the volume (V) of water is given as 95.0 mL.
To find the mass of water (m_water), we need to convert the volume to grams using its density:
Density of water = 1 g/mL
m_water = V_water * Density_water
= 95.0 mL * 1 g/mL
= 95.0 g
Now we can substitute the values into our equation:
Q_water = m_water * C_water * ΔT_water
= 95.0 g * 4.18 J/g⋅∘C * (-0.7∘C)
= -287.63 J
Since the steel rod loses the same amount of heat as the water gains, the heat lost by the steel rod will be:
Q_steel = -287.63 J
Now let's calculate the mass of the steel rod using the formula:
Q_steel = m_steel * C_steel * ΔT_steel
Since the steel rod is initially at 2.0∘C and the final temperature is 21.3∘C, we have:
ΔT_steel = 21.3∘C - 2.0∘C
= 19.3∘C
We know that the specific heat capacity of steel (C_steel) is 0.452 J/g⋅∘C.
Let's substitute the values into the equation:
Q_steel = m_steel * C_steel * ΔT_steel
= -287.63 J
Solve for m_steel:
m_steel = Q_steel / (C_steel * ΔT_steel)
= -287.63 J / (0.452 J/g⋅∘C * 19.3∘C)
≈ -8.85 g
The negative sign indicates that the steel rod lost heat to the water.
However, it's important to note that the calculated mass is negative, which implies that a negative amount of steel was present. This is not physically possible. Either there is an error in the given data, calculations, or assumptions made. Please check the problem statement and data provided.