Question

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)

Answer 1

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

*** 2 ***
molar heat capacity is J/mole C not J/gC

4.18 J/gC x (18.0 g / mole) = 75.2 J/moleC

Answer 2

mass of the steel is not 54.5 g.
I inputed the answer it didn't work. can you please help me

Answer 3

54.5 g steel doesnt work i inputed the answer it was incorrect

Answer 4

heat lost by water + heat gained by steel = 0
massH2O x specificheatH2O x (Tf-Ti) + masssteel x specificheatsteel x (Tf-Ti) = 0
100 x 4.18 x (21.2-22) + masssteel x 0.452 x (21.2-2) = 0
Solve for mass steel. I get about 40 g but you need to solve it exactly. Watch the signs. Post your work if you get stuck. Tf is final T. Ti is initial T.

Answer 5

the answer is 42.4

Answer 6

It would 24.9 grams
(105.5g)(0.5°C)(4.184J)=x(19.5)(0.452)

Answer 7

The answer with 3 sig figs and units is 43.6 g

Answer 8

38.9

Answer 9

I pooped my pants :(

Answer 10

it will be 33.5g

Answer 11

43.6 its 100 g

Answer 12

34.3

Answer 13

the correct answer 23.7 g

Answer 14

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.