Question #b4e85

1 Answer
Jul 12, 2015

Answer:

The initial temperature was #20.2""^@"C"#.

Explanation:

Once again, you need to know the specific heat of nickel before doing any calculations.

The specific heat of nickel is listed as being

#c_"nickel" = 0.44"J"/("g" ^@"C")#

In order to find the initial temperature of the sample, you'll use this equation

#q = m * c * DeltaT#, where

#q# - the amount of energy in the form of heat absorbed/released;
#m# - the mass of the sample;
#c# - the specific heat of the substance;
#DeltaT# - the change in temperature, defined as #T_"final" - T_"initial"#.

This time, you know that you supplied a certain amount of heat to the sample, 82.9 J to be precise, and that its final temperature was measured at #35.7^@"C"#.

You can use the values given to you to dolve for the initial temperature of the sample by

#q = m * c * underbrace(DeltaT)_(color(blue)(T_text(final) - T_"initial"))#

#82.9cancel("J") = 12.2cancel("g") * 0.44cancel("J")/(cancel("g") cancel(""^@"C")) * (35.7 - T_"initial")cancel(""^@"C")#

#82.9 = 191.6 - 5.368 * T_"initial"#

#T_"initial" = (191.6 - 82.9)/5.368 = 20.249""^@"C"#

Rounded to three sig figs, the answer will be

#T_"initial" = color(green)(20.2""^@"C")#