Question

How much heat in joules must be added to 0.250 mol of ar?

Answer 1

Answer:

Answer:

Q

=

50

,

800

J

=

12

,

100

cal

Explanation:

Use the equation that relates mass, specific heat, and change in temperature to the amount of heat of a substance:

Q

=

m

c

Δ

t

*Note that this equation assumes that temperature has units of Kelvin. We do not need to convert from Celsius to Kelvin because change in Celsius is the same as change in Kelvin.

Q

=

(

75.0

g

)

(

0.449

J

g

⋅

o

C

)

(

1535

o

C

−

25

o

C

)

Q

=

50

,

849.25

J

→

(Significant figures)  

Q

=

50

,

800

J

Q

=

50

,

849.25

J

⋅

1

calorie

4.186

J

=

12

,

147.46

cal

→

(significant figures)  

Q

=

12

,

100

cal

Explanation:

Answer 2

This is an incomplete question, here is a complete question.

How much heat in joules must be added to 0.250 mole of Ar to raise its temperature from 20 °C to 36 °C at constant pressure?

The heat added must be, 83.14 J

Explanation :

Formula used :

$q=nC_p\Delta T\\\\q=n\times C_p\times (T_2-T_1)$

where,

q = heat added = ?

n = number of moles of Ar = 0.250 mole

$C_p$ = heat at constant pressure for monoatomic gas (Ar) = $\frac{5}{2}R$

$T_1$ = initial temperature = $20^oC=273+20=293K$

$T_2$ = final temperature = $36^oC=273+36=309K$

R = gas constant = $8.314J/mol.K$

$q=n\times C_p\times (T_2-T_1)$

$q=0.250mol\times (\frac{5}{2}R)\times (309-293)K$

$q=0.250mol\times (\frac{5}{2}\times (8.314J/mol.K))\times (309-293)K$

$q=83.14J$

Thus, the heat added must be, 83.14 J

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