Question

Please help on the image below.

Answer 1

Heat capacity of Calorimeter = 10.7 J/°c

Further explanation

Heat lost=Heat gained

Q in = Q out

-Q lost(hot water)=Q gained (cold water+calorimeter)

-m.c.Δt=m.c.Δt+C.Δt

`\tt \rightarrow -50* 4.18* (32.7-43)=50* 4.18* (32.7-22.9)+(32.7-22.9)* C_(cal)\\\\\rightarrow 2152.7=2048.2+9.8* C_(cal)\\\\104.5~J=9.8* C_(cal)\\\\C_(cal)=(104.5)/(9.8)=10.7~J/^oC`