Part A A volume of 125 mL of H2O is initially at room temperature (22.00 C). A c

Question

Part A A volume of 125 mL of H2O is initially at room temperature (22.00 C). A chilled steel rod at 2.00 ° C is placed in the water. If the final temperature of the system is 21.00 , what is the mass of the steel bar? Use the following values specific heat of water = 4.18 J/(g·°C) specific heat of steel 0.452 J/(g. °C) Express your answer to three significant figures and include the appropriate units. Hints ? mass of the steel= 42.6 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining

Explanation / Answer

Let us denote rod by symbol 1 and water by symbol 2

m1 = to be calculated

T1 = 2.0 oC

C1 = 0.452 J/goC

m2 = 125.0 g

T2 = 22.0 oC

C2 = 4.18 J/goC

T = 21.0 J/goC

we have below equation to be used:

heat lost by 2 = heat gained by 1

m2*C2*(T2-T) = m1*C1*(T-T1)

125.0*4.18*(22.0-21.0) = m1*0.452*(21.0-2.0)

m1= 60.8 g

Answer: 60.8 g

Feel free to comment below if you have any doubts or if this answer do not work

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