Answer to Question #97443 in General Chemistry for Sarah

Question

A piece of stainless steel (specific heat = 0.50 J g-1°C-1) is transferred from an oven at (1.9x10^2) °C into 150. ml of water at (2.320x10^1) °C. The water temperature rises to 31.5 °C. Assuming that there is no heat loss and no loss of water vapour, calculate the mass (in g) of the piece of steel.

Answer 1

c(ss)=0.50 J*g-1*C-1

t1=190C

c(w)=4.22 J*g-1*C-1

t2=23.2C

m(w)=150 g

t3=31.5C

Q(ss)=Q(w)

Q(ss)=c(ss)*m(ss)*(t1-t3)

Q(w)=c(w)*m(w)*(t3-t2)

c(ss)*m(ss)*(t1-t3)= c(w)*m(w)*(t3-t2)

m(ss)= c(w)*m(w)*(t3-t2)/[c(ss)*(t1-t3)]= 4.22*150*(31.5-23.1)/[0.50*(190-31.5)]=67.1 g