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A coffee cup calorimeter contains 100.0 g of water at 10.0 ◦C. A 72.4 g sample of iron is heated to 100 ◦C and dropped into the calorimeter. Assuming there is no heat lost to the coffee cup, calculate the final temperature of the system. The specific heats of iron and water are 0.449 J/g ◦C and 4.184 J/g ◦C, respectively.

Answer :

Answer:

The final temperature is 16.47°C.

Explanation:

Water

m₁= 100 g

T₁= 10 °C

Cp₁= 4.184 J/g°C

Iron

m₂= 72.4 g

T₂=100 °C

Cp₂= 0.449 J/g°C

The temperature of iron i higher than the water that is why iron will loose and water will gain the heat.

We know that

Q= m Cp ΔT

Lets take final temperature is T

m₁ Cp₁ ( T- T₁ ) = m₂ Cp₂ ( T₂ -T)

100 x 4.184 ( T- 10) = 72.4 x 0.449 ( 100 - T)

12.87 ( T- 10) = 100 - T

12.87 T - 128.7 = 100 - T

13.87 T = 128.7 + 100

T=16.47°C

The final temperature is 16.47°C.

Answer:

16.5 º C

Explanation

heat lost by iron = heat gained by water

(72.4 g)(0.449 J/g/deg)(100 - T f) = (100  g)(4.184 J/g/deg)(T f - 10)

32.51(100-T f) = 418.4(T f-10)

3250 - 32.5 T f = 418.4 T f - 4184

7434 = 450.9 T f

T f = 16.48 degrees

Answer = 16.5 º C

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