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At an elevation where the boiling point of water is 93°C, 1.00 kg of water at 30°C absorbs 290.0 kJ from a mountain climber’s stove. Is this amount of thermal energy sufficient to heat the water to its boiling point? [cp of water = 4.18 J/(g · °C)] need more information to calculate can not be calculated even with more information no yes

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dsdrajlin

Answer:

Yes, it will be enough.

Explanation:

We can calculate the heat (Q) required to heat 1.00 kg of water from 30°C to 93°C using the following expression.

Q = cp × m × ΔT

where,

  • cp: specific heat capacity of water
  • m: mass of water
  • ΔT: change in the temperature

Q = cp × m × ΔT

Q = 4.18 J/g°C × 1.00 × 10³ g × (93°C-30°C)

Q = 263 kJ

Since 263 kJ are necessary, 290.0 kJ will be enough to heat the water.

pstnonsonjoku

The energy is sufficient to raise the temperature of the water to its boiling point.

We have the following information from the question;

Boiling point of water =  93°C

Mass of water = 1.00 kg or 1000 g

Heat capacity of water = 4.18 J/g · °C

Heat absorbed by water = 290.0 kJ or 290000 J

Initial temperature of the water = 30°C

Using the formula;

ΔH = mcθ

ΔH = Heat absorbed by the water

m = mass of the water

c = heat capacity of the water

θ = temperature rise (T2 - T1)

Substituting values;

290000 J = 1000 g × 4.18 J/g · °C (T2 - 30°C)

290000  = 4180T2 - 125400

T2 = 290000 + 125400/4180

T2 = 99.3°C

The energy is sufficient to raise the temperature of the water to its boiling point.

Learn more: https://brainly.com/question/13155407

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