Answer:
22.75° C
Explanation:
Heat loss = Heat gain
Thus, Q{tungsten}=Q{water}+Q{calorimter}Q
tungsten= Q(water) + Q(calorimeter) ------------(i)
At equilibrium final temperature of the system is the same for all it's constituents. Let this temperature be T{f}° C
Q=ms\∆T =c\∆TQ=msΔT=cΔT
m = mass of the substance (in gms.gms. )
s = specific heat capacity (in J/g°CJ/g°C )
ΔT= change in temperature of the substance ( °C )
c= heat capacity of the substance ( J/°CJ/°C )
Assuming we were given tha
m{tungsten}=19.40 gms.m
Temperature of Tungsten=97.35°C
m{water}=83.87gms
Temperature of water= 20.58°C = temperature of calorimeter
=20.58°C
At first the calorimeter will be at equilibrium with the water
Specific heat capacity{calorimeter}=1.83 J/°Cc
calorimeter
specific heat capacity{water}=1J/g°Cs
specific heat capacity{tungsten}=0.13J/g°C
Slotting in the values into (i), we have;
19.40 ×0.13(97.35-T{final} )=83.87×1(T{final} - 19.40 × 0.13(97.35−T{final}
=83.87×1(T{final}) −20.58)+1.83×(T{final} -20.58)20.58)+1.83∗(T{final}} -20.58)
2.522(97.35-T{final})=85.7(T{final}-20.58)2.522(97.35−T{final}
=85.7(T{final} −20.58)
2.522×97.35-2.522T{final} =85.7T{final}×85.7× 20.582.522×97.35−2.522T{final} =85.7T{final}−85.7×20.58
245.517+1763.706=88.222T{final} 245.517+1763.706=88.222T{final}
T{final}=2009.223/88.222T
f
=2009.223/88.222
=22.775° C
