contestada

a 5.00-g sample of aluminum pellets (specific heat capacity 0.89 j/c g) and a 10.00-g sample of iron pellets (specific heat capacity 0.45 j/c g) are heated to 100.0c. the mixture of hot iron and aluminum is then dropped into 97.3 g of water at 22.0c. calculate the final temperature of the metal and wa- ter mixture, assuming no heat loss to the surroundings. socratic.org

Respuesta :

The final temperature of the metal and wa- ter mixture, assuming no heat loss to the surrounding is 23.67 degree Celesius.

So, 5 grams of Aluminum and 10 grams of Iron pellets are heated to 100°C and added to 97.3 grams of water.

Final temperature = ?

All the heat given out by the aluminum and iron is absorbed by the water.

=> Let final Temperature be T.

For Heat Gained by Water:

T1 = 22 °C

T = ?

△T = T-22 °C

m = 0.0973 kg

c = 4200 J kg-1 °C-1

Q = 0.0973 kg * 4200 J kg-1 °C-1 * (T-22)  °C

Q1 = 408.66 * (T-22) J

For Heat Lost By Aluminum:

T1 = 100 °C

T = ?

△T = 100-T °C

m = 0.005 kg

c = 890 J kg-1 °C-1

Q = 0.005 kg * 890 J kg-1 °C-1 * (100-T)  °C

Q2 = 4.45 * (100-T) J

For Heat Lost By Iron:

T1 = 100 °C

T = ?

△T = 100-T °C

m = 0.01 kg

c = 450 J kg-1 °C-1

Q = 0.01 kg * 450 J kg-1 °C-1 * (100-T)  °C

Q3 = 4.5 * (100-T) J

Now,

Q1 = Q2 + Q3

408.66 * (T-22) = 4.45 * (100-T) + 4.5 * (100-T)

408.66T - 8990.52 = 445 - 4.45T + 450 - 4.5T

408.66T - 8990.52 = 895 - 8.95T

408.66T + 8.95T = 895 + 8990.52

417.61 T = 9885.52

T = 9885.52/417.61

T = 23.67 °C

The final temperature of the metal and water mixture, assuming no heat loss to the surrounding is 23.67 degree Celsius.

To learn more about final temperature Please click on the given link:

https://brainly.com/question/22338034

#SPJ4

Otras preguntas