calculate final temparature

Question

platinum has a specific heat capacity of 0.13 j/gk. if 280j are aplied to a 30g sample of platinum with an initial temparatire of 280k, what is the final temparature?

Answer 1

Given the information here, we can use the equation q = msΔT to solve for the final temperature of the platinum sample, where q is the amount of energy transferred (in Joules); m is the mass of the sample (in grams); s is the specific heat capacity (in J/g.K); and ΔT is the change in temperature, or T_final - T_initial (in Kelvin) Substituting everything into the equation and rearranging to solve for the missing variable, T_final, we get:

 

q = msΔT

280 J = (30 g)(0.13 J/g.K)(T_final - 280 K)

T_final = 352 K

 

 

Answer 2

Q = mcΔT = mc(T_f - T_i)

 

Q/(mc) = T_f - T_i

 

T_f = T_i + Q/(mc) = (280 K) + (280 J) / [(30 g)(0.13 J / gK)] ≅ 352 K