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Advocard [28]
3 years ago
10

How much energy (in KJ) is needed to melt 15g of ice at -4oC and raise it to a temperature of 115oC? Round the closest number, a

nd include units
Chemistry
1 answer:
WARRIOR [948]3 years ago
4 0

Answer:

The energy needed is 45.7659 kJ

Explanation:

The given mass of the ice, m = 15 g

The initial temperature of the ice, T₁ = -4°C

The temperature

The final temperature, T₂ = 115°C

The specific heat capacity of ice, c₁ = 2.09 J/(g·°C)

The latent heat of ice, l₁ = 334 J/g

The heat capacity of water, c₂ = 4.184 J/(g·°C)

The latent heat of vaporization, l₂ = 2260 J/g

The specific heat of steam, c₃ = 2.02 J/g

Therefore, we get the energy needed, ΔQ, as follows;

ΔQ = m·c₁·(T₂ - Tₐ) + m·l₁ + m·c₂(Tₓ - Tₐ) + m·l₂ + m·c₃·(T₂ - Tₓ)

Where;

Tₐ = The melting point temperature of water = 0°C

Tₓ = The boiling point temperature of water = 100°C

∴ ΔQ = 15×2.09×(0 - (-4)) + 15 × 334 + 15×4.184×(100 - 0) + 15×2260 + 15×2.02×(115 - 100) = 45,765.9

The energy needed, ΔQ = 45,765.9 J = 45.7659 kJ.

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In a car engine gasoline is burning to create mechanical energy which of the following statements is true? A. Some energy is los
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Answer:

Some energy is lost as heat

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4 0
3 years ago
The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of
Mademuasel [1]

Answer:

\large \boxed{\textbf{609 kJ}}  

Explanation:

The formula for the heat absorbed is

q = mCΔT

Data:

m = 2.07 kg

T₁ = 23 °C

T₂ = 191 °C

C = 1.75 J·°C⁻¹g⁻¹

Calculations:

1. Convert kilograms to grams

2.07 kg = 2070 g

2. Calculate ΔT

ΔT = T₂ - T₁ = 191 - 23  = 168 °C

3. Calculate q

q = \text{2070 g} \times 1.75 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 168 \,^{\circ}\text{C} = \text{609 000 J} = \textbf{609 kJ}\\\text{The heat energy required is }\large \boxed{\textbf{609 kJ}}

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3 years ago
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