Chemistry heat equation

Question

Show how to determine (and then calculate) the final temperature (in °C) of a 215 g copper pan when 6175 calories of energy are added to the copper pan that is initially at 20.3 °C. (the specific heat of copper is 0.0923 cal/g °C)

Answer 1

q = mC∆T

q = heat = 6175 cal

m = mass = 215 g

C = specific heat = 0.0923 cal/g/deg

∆T = change in temperature = ?

6175 cal = (215 g)(0.0923 cal/g/deg)(∆T deg)

∆T deg = 6175 cal/(215 g)(0.0923 cal/g/deg)

∆T = 311 degrees (3 sig. figs). This is the CHANGE in temperature from the original temperature.

Final temperature = 311º + 20.3º = 331.3ºC (4 significant figures since addition requires use of decimal)

Answer 2

215 g x ( 0.0923 cal/g °C) = 19.845 cal/1 °C (are required to heat this sample 1 degree C).

6175 cal x (1 °C/19.845 cal) = 311.162 °C

Begin at 20.3 °C, add 311.162 °C, = 331.462 °C = 331 sig figged to three sig figs