Question

calculate the mass of titanium (0.125 cal.g * C) that requires 75.5 cal to heat the metal from 20.5 to 32.0

Answer 1

C — specific heat capacity = 0.125

Q — heat energy = 75.5

∆T — change in temperature = 32-20.5=11.5

m — mass of substance = ?

$Q=mC∆T$

$m = \cfrac {Q}{C∆T}=\cfrac {75.5}{0.125*11.5}= 52.52~g$

Answer: 52.52 g