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zubka84 [21]
2 years ago
10

A kilogram of aluminum metal and a kilogram of water are each warmed to 75 ∘C and placed in two identical insulated containers.

One hour later, the two containers are opened and the temperature of each substance is measured. The aluminum has cooled to 35 ∘C while the water has cooled only to 66 ∘C.
Chemistry
1 answer:
zimovet [89]2 years ago
7 0

Answer:

Explanation:

The difference is in the amount of heat required to increase their temperature by 1 °C

It takes 4.19 joules to increase the temperature of water by 1 °C compared to Aluminium which is 0.900 joules to raise the temperature by  1 °C. Consequently, Aluminum will cool easily compare to water which will retain some of it heat longer because it does not heat up quickly like Aluminum.

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Write the formula for calcium chloride. Explain how you determined this formula
hichkok12 [17]

Calcium forms an ion with a positive 2 charge and chlorine forms an ion with a negative one charg, so the formula is <span>CaC<span>l2</span></span>

Group 1 metals and group 2 metals form positive ions by losing 1 and 2 electrons respectively. Non-metals in group 17 gain 1, group 16 gain 2 and group 15 gain 3. Elements which lose electrons form positive ions while elements that gain electrons form negative ions. 

To write a formula, you must balance charges so the overall charge is zero. A simple way to do this is to swap the # of the ion's charge and make it the subscript of the other ion. However, leave off the number 1 and reduce to lowest whole number ratio.

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3 years ago
Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which
Drupady [299]

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:= \dfrac{5 \ kg }{1590 \ kg/m^3}

= 0.0031 m³

Suppose \beta is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT

In ( \dfrac{V_2}{V_1})  = \beta (T_2-T_1)

V_2 = V_1 \times exp (\beta (T_2-T_1))

V_2 = 0.0031 \ m^3  \times exp  (1.2 \times 10^{-3} \times 20)

V_2 = 0.003175 \ m^3

However; the workdone = -PdV

W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)

W = - 7.6 J

The heat energy Q = Δ h

Q = mC_p(T_2-T_1)

Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ

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Answer:

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Answer:

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