95.0 mL of H_2O is initially at room temperature (22 degree C). A chilled steel rod at 2 degree C is placed in the water. If the final temperature of the system is 21.3 degree C, what is the mass of the steel bar?
the sum of the heats gained is zero.
masssteel*cs*(21.3-2)+masswater*cw*(21.3-22)=0
solve for masssteel. YOu will have to look up the specific heat cs for steel.
8.814
To solve this problem, we can use the principle of conservation of energy, specifically the equation for heat transfer:
Q = mcΔT
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
In this case, we have two substances: water (H2O) and the steel rod. We can calculate the heat transferred for each substance and then set them equal to each other, since the total heat transferred is conserved.
Let's start by calculating the heat transferred for the water.
Q_water = m_water * c_water * ΔT_water
We know the initial temperature of the water is 22°C, and the final temperature is 21.3°C. Therefore, ΔT_water = 21.3°C - 22°C = -0.7°C.
The specific heat capacity of water (c_water) is approximately 4.18 J/g°C.
Now we plug in the values and solve for Q_water:
Q_water = m_water * 4.18 J/g°C * -0.7°C
We want the mass of the steel bar, so let's focus on the heat transferred to the water. The heat transferred from the steel rod to the water is equal to the heat lost by the steel rod. Therefore, the heat transferred to the water can be written as:
Q_steel = m_steel * c_steel * ΔT_steel
The initial temperature of the steel rod is 2°C, and the final temperature is 21.3°C. Therefore, ΔT_steel = 21.3°C - 2°C = 19.3°C.
The specific heat capacity of steel (c_steel) is approximately 0.5 J/g°C.
Now we plug in the values and solve for Q_steel:
Q_steel = m_steel * 0.5 J/g°C * 19.3°C
The total heat transferred is conserved, so:
Q_water = Q_steel
m_water * 4.18 J/g°C * -0.7°C = m_steel * 0.5 J/g°C * 19.3°C
Now we can solve the equation for the mass of the steel bar (m_steel):
m_steel = (m_water * 4.18 J/g°C * -0.7°C) / (0.5 J/g°C * 19.3°C)
Plugging in the given values:
m_steel = (95.0 mL * 1 g/mL * 4.18 J/g°C * -0.7°C) / (0.5 J/g°C * 19.3°C)
Calculating this expression will yield the mass of the steel rod.