105 of is initially at room temperature (22.0). A chilled steel rod at 2.0 is placed in the water. If the final temperature of the system is 21.3, what is the mass of the steel bar?
105 WHAT of WHAT is initially at .....?
It's supposed to be 105 mL of H2O.
SH of water: 4.18 J/g deg C
SH of steel: 0.452 J/g deg C
To solve this problem, 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 steel rod.
The amount of heat gained or lost can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat gained or lost
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, the water gains heat and the steel rod loses heat. So we can write the equation:
m_water * c_water * ΔT_water = -m_steel * c_steel * ΔT_steel
Since the specific heat capacity (c) of water is much higher than steel, we can assume that the specific heat capacities of the water and the steel are constant (c_water = 4.18 J/g°C and c_steel = 0.452 J/g°C).
The change in temperature for the water can be calculated as:
ΔT_water = final temperature of the system - initial temperature of the water
= 21.3°C - 22.0°C
= -0.7°C
The change in temperature for the steel rod can be calculated as:
ΔT_steel = final temperature of the system - initial temperature of the steel rod
= 21.3°C - 2.0°C
= 19.3°C
Plugging these values into the equation, we have:
m_water * 4.18 J/g°C * (-0.7°C) = -m_steel * 0.452 J/g°C * 19.3°C
Simplifying the equation:
-2.9266 * m_water = -8.7236 * m_steel
Dividing both sides of the equation by -8.7236 * m_steel:
m_water / m_steel = -2.9266 / -8.7236
m_water / m_steel = 1 / 2.985
m_water ≈ 0.335 * m_steel
Since the mass of the water is given as 105 g, we can substitute this value into the equation:
105 g = 0.335 * m_steel
Dividing both sides of the equation by 0.335:
m_steel = 105 g / 0.335
m_steel ≈ 313.43 g
Therefore, the mass of the steel bar is approximately 313.43 grams.