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Lapatulllka [165]
3 years ago
13

Find the initial temperature of 45.0g of water if its final Temperature after submerging a metal with the mass of 8.5g is 22.0°C

. The initial temperature of metal before submerging it to water is 82.0°C
Chemistry
1 answer:
Korolek [52]3 years ago
8 0

The initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

<em>"Your question is not complete, it seems to be missing the following information;"</em>

the specific heat capacity of the metal is 0.45 J/g⁰C.

The given parameters;

  • <em>mass of water, </em>m_w<em> = 45 g</em>
  • <em>final temperature of the water, </em>t_w<em> = 22 ⁰C</em>
  • <em>mass of the metal, m = 8.5 g</em>
  • <em>initial temperature of the metal, t = 82 ⁰C.</em>
  • <em>specific heat capacity of the metal, c = 0.45 J/g⁰C.</em>

The initial temperature of the water will be calculated by applying the principle of conservation of energy;

<em>heat gained by water = heat lost by metal</em>

Q_{water} = Q_{metal}

m_wc_w \Delta t_w = mc\Delta t

where;

c_w <em>is the specific heat capacity of the water = 4.184 J/g⁰C.</em>

<em />

<em>Substitute the given values;</em>

45 x 4.184 x (22 - t) = 8.5 x 0.45 x (85 - 22)

4142.16  - 188.28t = 240.98

188.28t  = 4142.16  - 240.98

188.28t  = 3901.18

t = \frac{3901.18}{188.28} \\\\t = 20.7 \ ^0 C

Thus, the initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

Learn more here:brainly.com/question/15345295

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2ca(s)+o2(g) → 2cao(s) δh∘rxn= -1269.8 kj; δs∘rxn= -364.6 j/k
Alinara [238K]

Gibbs free energy change for the reaction at 29°C.  is equal to -1378.93 KJ.

<h3>What is Gibbs's free energy?</h3>

Gibbs free energy can be described as the enthalpy of the system minus the product of the temperature and entropy.

If the chemical reaction can be carried out under constant temperature ΔT = 0:

ΔG = ΔH – TΔS

The above equation is known as the Gibbs-Helmholtz equation.

ΔG > 0 non-spontaneous and endergonic and ΔG < 0 spontaneous and exergonic, ΔG = 0 is representing equilibrium.

Given the ΔS = -364 J/K, ΔH = -1269.8 KJ, T = 29°C = 29 + 273 = 302 K

ΔG = - 1269 - 302 × 364

ΔG =  -1269 KJ - 109.93 KJ

ΔG =  - 1378.93 KJ

Learn more about Gibbs's free energy, here:

brainly.com/question/13318988

#SPJ1

Your question is incomplete, most probably the complete question was,

2Ca(s)+O₂(g) → 2CaO(s)

ΔH∘rxn= -1269.8 kJ; ΔS∘rxn= -364.6 J/K

For this problem, assume that all reactants and products are in their standard states.

Calculate the free energy change for the reaction at 29°C.

6 0
2 years ago
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