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3 years ago

135 Joules of heat are required to increase the temperature of 10.0 grams of an unknown metal from 25.0∘C to 40.0∘C:

Chemistry

1 answer:

Sever21 [200]3 years ago

8 0

Answer:

270 Joules

Explanation:

The specific heat capacity equation will be used for this question i.e.

Q = m. c. ΔT

Where; Q = Amount of heat

m = mass of substance

C = specific heat capacity of

substance

ΔT = change in temperature

(Final temp - initial temp)

However, for this unknown metal, we need to find the specific heat first by saying; C = Q / mΔT

Q= 135J, C=?, m= 10g, ΔT = (40-25 = 15°C)

C = 135 / 15 × 10

C = 135/150

C = 0.9 J/g°C

If the specific heat capacity of the unknown metal is 0.9 J/g°C, then at a mass of 20.0g, and a ΔT of 15°C, the amount of heat needed is:

Q = m. c. ΔT

Q = 20 × 0.9 × (40-25)

Q = 18 × 15

Q = 270J

270 Joules of heat is needed to increase the temperature of 20g of the metal from 25 - 40°C

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29. A gas has a volume of 1.75 L at -23°C and 150.0 kPa.

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The answer for the following mention bellow.

Therefore the final temperature of the gas is 260 k

Explanation:

Given:

Initial pressure ( $P_{1}$ ) = 150.0 kPa

Final pressure ( $P_{2}$ ) = 210.0 kPa

Initial volume ( $V_{1}$ ) = 1.75 L

Final volume ( $V_{2}$ ) = 1.30 L

Initial temperature ( $T_{1}$ ) = -23°C = 250 k

To find:

Final temperature ( $T_{2}$ )

We know;

According to the ideal gas equation;

P × V = n × R ×T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

We know;

$\frac{P*V}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ × $\frac{V_{1} }{V_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

( $P_{1}$ ) represents the initial pressure of the gas

( $P_{2}$ ) represents the final pressure of the gas

( $V_{1}$ ) represents the initial volume of the gas

( $V_{2}$ ) represents the final volume of the gas

( $T_{1}$ ) represents the initial temperature of the gas

( $T_{2}$ ) represents the final temperature of the gas

So;

$\frac{150 * 1.75}{210 * 1.30}$ = $\frac{260}{T_{2} }$

( $T_{2}$ ) =260 k

Therefore the final temperature of the gas is 260 k

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