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Dimas [21]
3 years ago
14

 How much energy is needed to raise the temperature of 125g of water from 25.0oC to 35.0oC?  The specific heat of water is 4.184

J/goC.
Chemistry
1 answer:
Anvisha [2.4K]3 years ago
6 0

Hello!

To find the amount of energy need to raise the temperature of 125 grams of water from 25.0° C to 35.0° C, we will need to use the formula: q = mcΔt.

In this formula, q is the heat absorbed, m is the mass, c is the specific heat, and Δt is the change in temperature, which is found by final temperature minus the initial temperature.

Firstly, we can find the change in temperature. We are given the initial temperature, which is 25.0° C and the final temperature, which is 35.0° C. It is found by subtract the final temperature from the initial temperature.

35.0° C - 25.0° C = 10.0° C

We are also given the specific heat and the grams of water. With that, we can substitute the given values into the equation and multiply.

q = 125 g × 4.184 J/g °C × 10.0° C

q = 523 J/°C × 10.0° C

q = 5230 J

Therefore, it will take 5230 joules (J) to raise the temperature of the water.

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Answer : The energy removed must be, -67.7 kJ

Solution :

The process involved in this problem are :

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The expression used will be:

\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]

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c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC

\Delta H_{vap} = enthalpy change for vaporization = 33.9kJ/mole=33900J/mole=\frac{33900J/mole}{78.11g/mole}J/g=434.0J/g

Molar mass of benzene = 78.11 g/mole

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

\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]

\Delta H=-67682.5J=-67.7kJ

Therefore, the energy removed must be, -67.7 kJ

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