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Feliz [49]
3 years ago
10

Based on the definition of specific heat and the information provide below, if the same amount of heat is added to 1 kg of all t

hese substances, which substance will raise in temperature to the highest value? Specific Heat Values (J/kg°C) of Different Materials: Aluminum: 897 Helium: 5193 Steel: 490 Water: 4182
Is it:
Aluminum
Helium
Steel
Water
Chemistry
2 answers:
REY [17]3 years ago
7 0
It would be the one with the highest Specific Heat constant

The equation for specific heat is:

Q=mCT

If we rearrange for T, we get:

T=Q/mC

Since m is just 1 Kg, we can ignore it and say that:

T=Q/C

Let’s just give an arbitrary number to C and solve for each individual species

Q=1 J

Helium - 1J/5193 = .00019257

Steel - 1J/490 = .00204082

Aluminum - 1J/897 =.00111483

Water - 1J/4182 = .00023912

This proves that Steel, with the lowest Specific Heat will increase the most.
goldfiish [28.3K]3 years ago
3 0

Answer:

The correct answer is steel.

Explanation:

The specific heat capacity refers to the concentration of energy needed to elevate the temperature of one kilogram of a substance by one degree. Thus, the greater the number, the more will be the energy required. Therefore, steel will rise in temperature to the highest value as less energy will be needed to elevate its temperature because of the lowest specific heat values of all the mentioned options.

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g The combustion of 1.877 1.877 g of glucose, C 6 H 12 O 6 ( s ) C6H12O6(s), in a bomb calorimeter with a heat capacity of 4.30
Mekhanik [1.2K]

Answer:

-2.80 × 10³ kJ/mol

Explanation:

According to the law of conservation of energy, the sum of the heat absorbed by the bomb calorimeter (Qcal) and the heat released by the combustion of the glucose (Qcomb) is zero.

Qcal + Qcomb = 0

Qcomb = - Qcal [1]

We can calculate the heat absorbed by the bomb calorimeter using the following expression.

Qcal = C × ΔT = 4.30 kJ/°C × (29.51°C - 22.71°C) = 29.2 kJ

where,

C: heat capacity of the calorimeter

ΔT: change in the temperature

From [1],

Qcomb = - Qcal = -29.2 kJ

The internal energy change (ΔU), for the combustion of 1.877 g of glucose (MW 180.16 g/mol) is:

ΔU = -29.2 kJ/1.877 g × 180.16 g/mol = -2.80 × 10³ kJ/mol

3 0
4 years ago
Marianne designs an experiment involving electrically charged objects. She wants to know which objects will be attracted to a ne
eduard

Answer:

OUTCOME VARIABLE: Attraction to negatively charged balloon

Explanation:

In an experiment, two major variables must be included viz: the independent variable and the dependent variable. The independent variable also referred to as the test variable is the variable that is being changed in the experiment while the dependent or outcome variable is the variable that responds to the changes made to the independent variable. The outcome variable is the variable that is measured in an experiment.

In this experiment involving Marianne wanting to know which objects will be attracted to a negatively charged balloon. The type of objects are the independent variable while the ATTRACTION is the outcome or dependent variable as it is dependent on the type of object used.

8 0
3 years ago
Pls help fast!! Test due soon!!!
blsea [12.9K]

Answer:

its compound option c compound comprises of two or more atoms

7 0
3 years ago
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A compound is composed of only C, H, and O. The combustion of a 0.519-g sample of the compound yields 1.24 g of CO 2 and 0.255 g
vredina [299]

Answer:

      \large\boxed{\large\boxed{C_3H_3O}}

Explanation:

<u>1. Calculate the mass of C in 1.24 g of CO₂</u>

   1.24gCO_2\times \dfrac{12.011gC}{44.01gCO_2}=0.3384gC

<u>2. Calculate the mass of H in 0.255g of H₂O</u>

     0.255gH_2O\times \dfrac{2.016gH}{18.015gH_2O}=0.02854gH

<u />

<u>3. Calculate the mass of O by difference</u>

     0.519g-0.3384g-0.02854g=0.15206g

<u>4. Convert every mass to number of moles:</u>

  • C: 0.3384g / (12.011g/mol) = 0.02817 mol
  • H: 0.02854g / (1.008g/mol) = 0.02831 mol
  • O: 0.15206g / (15.999g/mol) = 0.009504 mol

<u />

<u>5. Divide every number of moles by the least number of moles (0.009504):</u>

  • C: 0.02817/0.009504 = 2.96 ≈ 3
  • H: 0.02831 / 0.009504 ≈ 3
  • O: 0.009504 / 0.009504 = 1

<u>6. Those numbers are the respective subscripts in the empirical formula:</u>

     C_3H_3O

7 0
3 years ago
[Timed] Which type of reaction does this diagram represent?
IRINA_888 [86]
Nuclear fission because an atom is splitting....
5 0
4 years ago
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