What is the specific heat of aluminum if the temperature of a 30.0 g sample of a

Question

What is the specific heat of aluminum if the temperature of a 30.0 g sample of a aluminum is increased by 20 degree C when 100 J of heat is added? What is the specific heat of silicon if the temperature of a 15.0 g sample of silicon is increased by 138 degree C when 1100 J of heat id added? How much heat must be added to a 821 g sample of a gold to increase its temperature by 162 degree C? The specific heat of gold is 0.13 J/g degree C. If 40.5 J of heat is added to a 15.4 g sample of silver, how much will the temperature increase by? The specific heat is silver is 0.235 J/g degree C. When a 257. 0 degree g sample of a Nal dissolves in 180. 0 g of water in a calorimeter, the temperature rises from 40.5 degree C to 60.4 C. Calculate Delta H for the process. When a 169.0 g sample of NaOH dissolves in 700.0 g water in a calorimeter, the temperature rises from 22.4 degree C to 56.6 degree C. Calculate Delta H for the process. When a 128 g sample of KCl dissolves in 75.0 g of water in a calorimeter, the temperature drops from 51.0 degree C to 21.6 degree C. Calculate Delta H for the process

Explanation / Answer

Formula given is,

Q = m x C x T …………(1)

Where,

Q = Heat added/absorbed in J.

m = Mass of the substance in g

T = Rise in temperature in oC when Q amount of heat is added/absorbed.

C = Specific heat capacity in J/go C.

Bases on this given numericals are solved.

1) Given for Aluminium,

m = 30.0 g, Q = 100 J, T = 20 oC, and C = ?

Using eq. (1)

100 = 30.0 x C x 20

100 = 600 x C

C = 100/600

C = 1/6

C = 0.167 J/go C

Specific heat of Al is 0.167 J/go C.

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2) For Silicon,

m = 15.0 g, Q = 1100 J, T = 138 oC, and C = ?

Using eq. (1)

1100 = 15.0 x C x 138

1100 =2070 x C

C = 1100/2070

C = 0.531

C = 0.531 J/go C

Specific heat of Si is 0.531 J/go C.

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3) For Gold (Au),

m = 821 g, Q =?, T = 162 oC, and C = 0.13 J/g oC

Using eq. (1)

Q = 821 x 0.13 x 162

Q = 17290.26 J

Q = 17290 J (rounded)

C = 17.29 kJ

17.29 kJ heat must be added for desired result.

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4) For Silver (Ag),

m = 15.4 g, Q = 40.5 J, T =?, and C = 0.235 J/g oC

Using eq. (1)

40.5 = 15.4 x 0.235 x T

40.5 = 3.619 x T

T = 40.5 / 3.619

T = 11.2

T = 11.2 o C

Temperature of Ag will increase by 11.2 oC

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