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-Dominant- [34]

2 years ago

14

If the specific heat of water is 4. 18 j/g∙°c, how much heat is required to increase the temperature of 1. 20 kg of water from 2

3. 00 °c to 39. 00 °c?

1 answer:

Mkey [24]2 years ago

8 0

The heat required to increase the temperature of 1.20 g of water is 80,256 k/j

<h3>What is specific heat?</h3>

The amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

4. 18 J heat is required to change the temperature of 1.20 kg of water from 23. 00 °c to 39. 00 °c.

The heat required to raise temperature is the product of mass, specific heat and temperature change

1,200 × 4.18 × (39 − 23) = 80,256

Thus, the heat required to increase the temperature of 1.20 g of water is 80,256 k/j

Learn more about specific heat

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The Density of pure carbon in Diamond form is 3.52 g/cm^3. How many cubic inches would 23.7 moles of pure diamond occupy?

Vedmedyk [2.9K]

Answer : The volume of pure diamond is $0.493inch^3$

Explanation : Given,

Density of pure carbon in diamond = $3.52g/cm^3$

Moles of pure diamond = 23.7 moles

Molar mass of carbon = 12 g/mol

First we have to calculate the mass of carbon or pure diamond.

$\text{ Mass of carbon}=\text{ Moles of carbon}\times \text{ Molar mass of carbon}$

Molar mass of carbon = 12 g/mol

$\text{ Mass of carbon}=(23.7moles)\times (12g/mole)=284.4g$

Now we have to calculate the volume of carbon or pure diamond.

Formula used:

$Density=\frac{Mass}{Volume}$

Now putting all the given values in this formula, we get:

$3.52g/cm^3=\frac{284.4g}{Volume}$

Volume = $80.8cm^3$

As we know that:

$1cm^3=0.061inch^3$

So,

Volume = $0.061\times 80.8inch^3$

Volume = $0.493inch^3$

Therefore, the volume of pure diamond is $0.493inch^3$

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3 years ago

¿Cuáles son las características de los materiales de laboratorio? Por ejemplo: exactitud, resistencia a la temperatura, etc.

boyakko [2]

Answer:

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3 years ago

What is filtration? ㅜㅜ ㅜㅜ

Aleks04 [339]

Answer:

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Example:

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3 years ago

A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

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3 years ago

After the solution reaches equilibrium, what concentration of zn2 (aq remains?

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According to sources, the most probable answer to this query is that when solutions reaches equilibrium, the amount of concentration of two or more matter combined in this solution becomes equal.

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