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mr Goodwill [35]

3 years ago

6

A 17.98-g piece of iron absorbs 2056.5 joules of heat energy, and its temperature changes from 25°C to 200°C. Calculate the spec

ific heat capacity of iron.

1 answer:

V125BC [204]3 years ago

7 0

Answer: $0.65\ J/g.^{\circ}C$

Explanation:

Given

Mass of iron piece is $m=17.98\ g$

Heat absorbed $Q=2056.5\ J$

Temperature changes from $25^{\circ}C\ to\ 200^{\circ}C$ i.e.

$\Delta T=200-25=175^{\circ}C$

Heat absorbed is given by $Q=mc\Delta T\quad [c=\text{specific heat of material}]$

Insert the values

$\Rightarrow 2056.5=17.98\times c\times 175\\\\\Rightarrow c=\dfrac{2056.5}{3146.5}\\\\\Rightarrow c=0.65\ J/g.^{\circ}C$

Thus, the specific heat of iron is $0.65\ J/g.^{\circ}C$

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4 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

Now put all the given values in the above formula, we get:

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

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