to heat 10.2 g of silver from 112∘C to 255 ∘C

Question

Answer in joules and then calories.

Answer 1

spec heat = J/gC

Spec heat of silver is .233J/gC

J = c_p  x  g  x  ΔC

J = .233 J/gC  x  10.2g  x 143

J = 340.J

1J/0.000239006 Kcal

340. J x (0.000239006 Kcal/1J)  x  (1000cal/1Kcal) = 81.3 cal

 

Answer 2

Q = mcΔT

 

m = 10.2 g = 0.0102 kg

c = specific heat of silver [Look it up in your book, preferably in units of J/(kgºC).]

ΔT = (255 - 112)ºC = ? ºC

 

Plug in the numbers and get Q in joules, then convert joules to calories (look up the conversion factor in your book).