Chemistry Problem, Help Please

Question

A 52.00 g sample of a substance is initially at 29.9 °C. After absorbing 1863 J of heat, the temperature of the substance is 185.6 °C. What is the specific heat (c) of the substance?

Answer 1

Justis,

Just use the same formula again, but solve for c instead of Q:

Q = cMΔT   then  c = Q/(MΔT)

where
Q = heat energy needed for that material to get desired temperature change (in Joules)
M = mass (in grams) so you have to convert from kilograms.
c = specific heat constant for the material being heated [in /(grams ^oC)]
ΔT = change in temperature (in ^oC)

 

So   c = 1863joules/[(52 grams)(185.6 ^oC - 29.9 ^oC)]  = 1863/[52(155.7)] = 1863/8096.4

       c = 0.230 /g^oC 

Q = (1.75)(2.17 x 1000)(191 - 23) = (1.75)(2170)(168) = 637,980 Joules