Heat of Fusion How much thermal energy is needed change 50.0 g of ice at -20.0°C to water at 10.0°C?

Answers

Answer 1

Answer: We should divide the problem into 3 separate processes.

1) Bring the temperature of the ice from $-20.0^(\circ)C$ to its melting point ( $0^(\circ)C$ ): the amount of heat needed in this process is
$Q_1=mC_s \Delta T$
where
$m=50.0 g$ is the mass of the ice
$C_s = 2.108 J/g^(\circ)C$ is the specific heat capacity of ice
$\Delta T=0^(\circ)C-(-20^(\circ)C)=20^(\circ)C$ is the increase of temperature

Plugging numbers into the equation, we find
$Q_1 = (50.0 g)(2.108 J/g^(\circ)C)(20^(\circ)C)=2108 J$

2) Fusion of ice
When the ice is at melting point, we need to add a certain amount of heat in order to melt it, and this amount of it is given by:
$Q_2 = mL_f$
where
$m=50.0 g$ is the mass of ice
$L_f = 334 J/g$ is the latent heat of fusion of ice

Plugging numbers into the equation, we find
$Q_2 = mL_f = (50.0g)(334 J/g)=16700 J$
During this phase transition, the temperature of the ice/water does not change.

3) Bring the temperature of the water from $0^(\circ)C$ to $10^(\circ)C$

The amount of heat needed for this process is
$Q_3 = mC_s \Delta T$
where
$m=50.0 g$ is the mass of water
$C_s = 4.187 J/g^(\circ)C$ is the specific heat capacity of water
$\Delta T=10^(\circ)C-0^(\circ)C=10^(\circ)C$ is the increase of temperature

Plugging numbers into the equation, we find
$Q_3 = (50.0 g)(4.187 J/g^(\circ)C)(10.0^(\circ)C)=2094 J$

--> therefore, the total energy needed for the whole process is:
$Q=Q_1+Q_2+Q_3=2108 J+16700 J+2094 J=20902 J=20.9 kJ$

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Answers

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Answers

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Answers

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By definition, the kinetic energy is given by:

Where,

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Where,

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Answer:

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Answers

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Explanation:

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Answers

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hope this helps! Giving me brainliest answer would be appreciated

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Answers

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