A61.53 gram sample of iron (with a heat capacity of 0.450 Vg\'C) is heated to 10

Question

A61.53 gram sample of iron (with a heat capacity of 0.450 Vg'C) is heated to 100.00 It is then transferred to a coffee cup calorimeter containing 52.77 g of water (specific heat of 4.184V g'C) initially at 20.16 C. If the final temperature of the system is 28.93, what wes the heat gained by the calorimeter? If the calorimeter had a mass of 26.91 g what is the heat capacity of the calorimeter? J absorbed by the calorimeter Points avalable on this attempt: 1.2 oforiginal 2 Unlimited aremps. Score on last attempt: (O·0), Score in gradebook: (0, 0). Out of (1, 1)

Explanation / Answer

1)

step 1: calculate heat given by iron

we have:

m = 61.53 g

C = 0.45 J/g.oC

Ti = 100 oC

Tf = 26.91 oC

we have below equation to be used:

Q = m*C*(Tf-Ti)

Q = 61.53*0.45*(26.91-100.0)

Q = -2024 J

negative sign just shows that heat is released

step 2: calculate heat absorbed by water

we have:

m = 52.77 g

C = 4.184 J/g.oC

Ti = 20.16 oC

Tf = 26.91 oC

we have below equation to be used:

Q = m*C*(Tf-Ti)

Q = 52.77*4.184*(26.91-20.16)

Q = 1490 J

Step 3:

remaining heat must be absorbed by calorimeter

Q = 2024 J - 1490 J

= 534 J

Answer: 534 J

2)

for calorimeter:

we have:

Q = 534 J

m = 26.91 g

Ti = 20.16 oC

Tf = 26.91 oC

we have below equation to be used:

Q = m*C*(Tf-Ti)

534.0 = 26.91*C*(26.91-20.16)

C = 2.94 J/g.oC

Answer: 2.94 J/g.oC

Feel free to comment below if you have any doubts or if this answer do not work. I will correct it and submit again if you let me know

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