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3 years ago

10. What is the specific heat of an unknown substance if a 2.50 g sample releases 12 calories as its

Chemistry

1 answer:

Sergeu [11.5K]3 years ago

4 0

Answer:

c = 4016.64 j/g.°C

Explanation:

Given data:

Mass of substance = 2.50 g

Calories release = 12 cal (12 ×4184 = 50208 j)

Initial temperature = 25°C

Final temperature = 20°C

Specific heat of substance = ?

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

Solution:

Q = m.c. ΔT

ΔT = T2 - T1

ΔT = 20°C - 25°C

ΔT = -5°C

50208 j = 2.50 g . c. -5°C

50208 j = -12.5 g.°C .c

50208 j /-12.5 g.°C = c

c = 4016.64 j/g.°C

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A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample star

Trava [24]

Answer:

The specific heat capacity of quartz is 0.71 J/g°C.

Explanation:

Heat lost by quartz will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of quartz= $m_1=51.9 g$

Specific heat capacity of quartz= $c_1=?$

Initial temperature of the quartz= $T_1=97.8^oC$

Final temperature = $T_2=T = 19.3^oC$

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= $m_2=300.0 g$

Specific heat capacity of water= $c_2=4.18 J/g^oC$

Initial temperature of the water = $T_3=17.0 ^oC$

Final temperature of water = $T_2=T=19.3^oC$

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(51.9 gc_1\times (19.3^oC-97.8^oC))=300.0 g\times 4.18 J/g^oC\times (19.3^oC-170^oC)$

we get:

$c_1 =0.7079 J/g^oC\approx 0.71 J/g^oC$

The specific heat capacity of quartz is 0.71 J/g°C.

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3 years ago

How many moles of solve (Ag) are equivalent to 68.3g Ag

Zina [86]

Moles Ag = 68.3 g / 107.868 g/mol=0.633

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4 years ago

A flask of fixed volume contains 1.00 mole of gaseous carbon dioxide and 88.0 g of solid carbon dioxide. the original pressure a

Lynna [10]

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4 years ago

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What kind of graph would best show the info in the data table below?

Rasek [7]

Answer:a line or a circle but i think circle

Explanation:

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3 years ago

Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell

Assoli18 [71]

The question is incomplete, here is the complete question:

Solid cesium iodide has the same kind of crystal structure as CsCl which is pictured below: If the edge length of the unit cell is 456.2 pm, what is the density of CsI in $g/cm^3$

The image is attached below.

<u>Answer:</u> The density of CsI is $9.09g/cm^3$

<u>Explanation:</u>

To calculate the density of metal, we use the equation:

$\rho=\frac{Z\times M}{N_{A}\times a^{3}}$

where,

$\rho$ = density

Z = number of atom in unit cell = 2 (BCC)

M = atomic mass of CsI = 259.8 g/mol

$N_{A}$ = Avogadro's number = $6.022\times 10^{23}$

a = edge length of unit cell = $456.2pm=456.2\times 10^{-10}cm$ (Conversion factor: $1cm=10^{10}pm$ )

Putting values in above equation, we get:

$\rho=\frac{1\times 259.8}{6.022\times 10^{23}\times (456.2\times 10^{-10})^3}\\\\\rho=9.09g/cm^3$

Hence, the density of CsI is $9.09g/cm^3$

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3 years ago