100 mL of is initially at room temperature (22)C. A chilled steel rod at 2 C is placed in the water. If the final temperature of the system is 21.2 C, what is the mass of the steel bar?
Specific heat of water = 4.18J/g(C)
Specific heat of steel = 0.452J/g(C)
I pooped my pants :(
38.9
mass of the steel is not 54.5 g.
I inputed the answer it didn't work. can you please help me
54.5 g steel doesnt work i inputed the answer it was incorrect
heat lost by water = heat gained by steel
(m Cp dT) water = (m Cp dT) steel
mass steel = (m Cp dT) water / (Cp dT) steel
mass steel = (125 g) x (4.18 J/gC) x (22.0-21.1 C) / [(0.452 J/gC) x (21.1 - 2.0 C)] = 54.5 g steel
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molar heat capacity is J/mole C not J/gC
4.18 J/gC x (18.0 g / mole) = 75.2 J/moleC
43.6 its 100 g
34.3
the correct answer 23.7 g
To calculate the mass of the steel bar, we can use the equation:
(mass of water)(specific heat of water)(change in temperature of water) = (mass of steel)(specific heat of steel)(change in temperature of steel)
Let's plug in the given values:
(100 g)(4.18 J/g°C)(21.2°C - 22°C) = (mass of steel)(0.452 J/g°C)(21.2°C - 2°C)
Simplifying:
(-83.6 J) = (mass of steel)(-0.452 J/°C)(19.2°C)
Dividing both sides by (-0.452 J/°C)(19.2°C):
mass of steel = (-83.6 J) / [(-0.452 J/°C)(19.2°C)]
mass of steel = 236.73 g
Therefore, the mass of the steel bar is approximately 236.73 g.
It would 24.9 grams
(105.5g)(0.5°C)(4.184J)=x(19.5)(0.452)