95.0mL of H2O is initially at room temperature (22.0?C). A chilled steel rod at

Question

95.0mL of H2O is initially at room temperature (22.0?C). A chilled steel rod at 2.0?C is placed in the water. If the final temperature of the system is 21.5?C, what is the mass of the steel bar? Specific heat of water = 4.18 J/g??C Specific heat of steel = 0.452 J/g??C Express your answer numerically, in grams, to one significant figure.

i got 22.5 and it was wrong because either the rounding or the # if sig figs is wrong. I dont know how its going to be just 1 sig figs. I tried 23 to make it a whole number but it was still wrong.

Explanation / Answer

taking reference temperature as 2degC,

95*4.18*(22-2) + m*0.452*(2-2) = 95*4.18*(21.5-2) + m*0.452*(21.5-2)

=> m = 22.526 g = 22.53 g (in 2 significant figure) = 22.5 g (in 1 significant figure)

as par the above data, your answer is correct. Just check if you copied a wrong data !

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