Question

Calculate the final temperature of the system: A 50.0 gram sample of water initially at 100 °C and a 100 gram sample initially at 27.32 °C are mixed. The specific heat of water is 4.184 J/gC). Record your answer in scientific notation using 3 significant figures.

Answer 1

Answer:

The final temperature of the water system is 51.5 °C.

Explanation:

A 50.0 gram sample of water initially at 100 °C

Mass of the water = 50 g

Initial temperature of the water = $T_i=100$

Final temperature of the water after mixing = $T_f=$

-Q' = heat lost by 50.0 g of water

$-Q'=mc\Delta (T_f-T_i)$

A 100.0 gram sample of water initially at 27.32°C

Mass of the water ,m'= 10.00 g

Initial temperature of the water = $T_i'=100$

Final temperature of the water after mixing = $T_f=$

Q = heat gained by 100.0 g of water after mixing

$Q=m'c\Delta (T_f-T_i')$

-Q'=Q (Energy remains conserved)

$-(mc\Delta (T_f-T_i))=m'c\Delta (T_f-T_i')$

$-(50 g\times 4.184J/g^oC(T_f-100^oC))=100 g\times 4.184J/g^oC(T_f-27.32^oC)$

$T_f=51.54^oC\approx 51.5^oC$

The final temperature of the water system is 51.5 °C.