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barxatty [35]
3 years ago
5

Show the calculation of the final temperature for a 20.8 gram piece of iron heated to 100oC which has been added to a 55.3 gram

sample of water at 25.3oC in a coffee cup calorimeter.
c (water) = 4.184 J/g oC; c (Fe) = 0.449 J/g oC
Chemistry
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

Final Temperature = 28.2 oC

Explanation:

Information given;

Mass of Iron = 20.8g

Initial Temperature of Iron = 100C

Mass of water = 55.3g

Initial temperature of water = 25.3 C

The presence of a coffee cup calorimeter hints that there is no heat loss to the surrounding and that the iron and water are at thermal equilibrium.

Thermal equilibrium means that there is no heat transfer going on between the bodies, which simply means that the bodies are at the same temperature.

Hence, both bodies would the same final temperature (T2)

H = M * C * ΔT (For iron)

H = 20.8 * 0.449 * ( 100 - T2)

H = 9.3392 ( 100 - T2)

H = 933.92 - 9.3392T2

H = M * C * ΔT (For water)

H = 55.3 * 4.184 * (T2 - 25.3)

H = 231.3752 (T2 - 25.3)

H = 231.3752T2 - 5853.79

Since they are in thermal equilibrium it means H (Iron) = H (water).

This leads to;

933.92 - 9.3392T2 = 231.3752T2 - 5853.79

231.3752T2 +  9.3392T2  = 5853.79 + 933.92

240.7144 T2 = 6787.71

T2 = 28.2 oC

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Galileo discovered that objects that are more dense, or have more mass, fall at a faster rate than less dense objects, due to this air resistance. A feather and a rubber ball dropped together. Air resistance causes the feather to fall more slowly, while the ball falls more fast.

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Which elements are more likely to lose electrons?<br> Ca, Li, F and Ne
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Ca, they only have two valence electrons, in order to become more stable, they would like to lose all of them
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What technology can make coal a cleaner fuel? fluidized-bed combustion building smoke stacks taller implementing a prewash of co
anastassius [24]
Answer : The technology that can make coal a cleaner fuel is to use lower quality of coal with high sulfur content.

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4 0
3 years ago
4.45 kcal of heat was added to increase the temperature of a sample of water from 23.0 °C to 57.8 °C. Calculate
Alona [7]

Answer:

m = 4450 g

Explanation:

Given data:

Amount of heat added = 4.45 Kcal ( 4.45 kcal ×1000 cal/ 1kcal = 4450 cal)

Initial temperature = 23.0°C

Final temperature = 57.8°C

Specific heat capacity of water = 1 cal/g.°C

Mass of water in gram = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 57.8°C - 23.0°C

ΔT = 34.8°C

4450 cal = m × 1 cal/g.°C × 34.8°C

m = 4450 cal / 1 cal/g

m = 4450 g

4 0
3 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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