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4 years ago

How many joules of heat are removed from a 21.0 g sample of water if it is cooled from 34.0°C

Chemistry

1 answer:

yaroslaw [1]4 years ago

4 0

Answer:

527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, the sensible heat of a body is the amount of heat received or transferred by a body when it undergoes a change in temperature without there being a change in physical state (solid, liquid or gaseous). The equation that allows to calculate this heat exchange is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT=Tfinal-Tinitial is the change in temperature.

In this case:

c= 4.184 $\frac{J}{g*C}$
m=21 g
ΔT=Tfinal-Tinitial=28 °C - 34 °C=-6 °C

Replacing:

Q= 4.184 $\frac{J}{g*C}$ * 21 g* (-6 C)

Q= - 527.184 J

To lower the temperature, heat has to be given, for that the final temperature must be lower than the initial temperature; and it receives the name of transferred heat and has a negative value, as in this case.

<u><em> 527.184 J of heat is removed from a 21 g water sample if it is cooled from 34.0 ° C to 28.0 ° C.</em></u>

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4 years ago

The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is ________ kj. the specific heats of ice, wat

liubo4ka [24]

Answer : The enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

$(1):ice(-25^oC)\rightarrow ice(0^oC)\\\\(2):ice(0^oC)\rightarrow water(0^oC)\\\\(3):water(0^oC)\rightarrow water(90^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{ice}\times (T_2-T_1)]+\Delta H_{fusion}+[m\times c_{water}\times (T_3-T_2)]$

where,

$\Delta H$ = enthalpy change

m = mass of water = $1mole\times 18g/mole=18g$

$c_{ice}$ = specific heat of ice = 2.09 J/gk

$c_{water}$ = specific heat of water = 4.18 J/gk

$\Delta H_{fusion}$ = enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion : $0^oC=273k$

$T_1$ = initial temperature of ice = $0^oC=273k$

$T_2$ = final temperature of ice = $-25^oC=273+(-25)=248k$

$T_3$ = initial temperature of water = $0^oC=273k$

$T_4$ = final temperature of water = $90^oC=273+90=363k$

Now put all the given values in the above expression, we get

$\Delta H=[18g\times 2.09J/gK\times (273-248)k]+0.00601J+[18g\times 4.18J/gK\times (363-273)k]$

$\Delta H=7712.106J=7.712KJ$ (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at $-25.0^oC$ to water at $90^oC$ is, 7.712 KJ

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Explanation:

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It may be monoatomic , diatomic, triatomic etc.

For example,

Hydrogen molecule consist of two atoms H₂

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It may be combination of atoms of different elements.

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