Question

Using the accepted value for the specific heat of tungsten (See the References tool), calculate the final temperature of the water. Assume that no heat is lost to the surroundings.

Answer 1

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

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Temperature of Tungsten=97.35°C

m{water}=83.87gms

Temperature of water= 20.58°C = temperature of calorimeter

=20.58°C

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At first the calorimeter will be at equilibrium with the water

Specific heat capacity{calorimeter}=1.83 J/°Cc

calorimeter

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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}

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=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}

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=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}

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T{final}=2009.223/88.222T

f

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=2009.223/88.222

=22.775° C