Physics, 24.06.2019 11:00, dianepowers1

The questions are 1. how much heat is needed to melt 10.0kg of lead at it’s melting point? 2. how much heat is needed to vaporize 10.0 kg of mercury at its boiling point? 3. a temperature of a 0.25 kg sample of copper decreases from 75.0 degrees celsius to 25.0 degrees celsius. how much heat flows out of the copper sample? 4. when 8,000 joules of heat are removed from a sample of mercury, the temperature of the sample is decreased from 40 degrees celsius to 15 degrees celsius. what is the mass of the sample? 5. thermal energy of a substance is the sum of the potential energy and kinetic energy of the particles that make it up. let’s say you had a hot cup of water and a cold cup of water, then mixed them. illustrate the behavior of the particles before and after thermal equilibrium is reached

The questions are 1. how much heat is needed to melt 10.0kg of lead at it’s melting point? 2. how m

Answers: 2

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Answers

answered: DASASDAEDWEDA

1. $2.3\cdot 10^5 J$

The amount of heat needed to melt a substance at its melting point is:

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the latent heat of fusion of the substance

For the sample of lead in this problem,

m = 10.0 kg

$\lambda_f = 23000 J/kg$ (latent heat of fusion of lead)

Substituting,

$Q=(10.0 kg)(23000 J/kg)=2.3\cdot 10^5 J$

2. $2.96\cdot 10^6 J$

The amount of heat needed to evaporate a substance at its boiling point is:

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the latent heat of vaporization of the substance

For the sample of mercury in this problem,

m = 10.0 kg

$\lambda_v = 2.96\cdot 10^5 J/kg$ (latent heat of vaporization of mercury)

Substituting,

$Q=(10.0 kg)(2.96\cdot 10^5 J/kg)=2.96\cdot 10^6 J$

3. -4875 J

The amount of heat released from the sample of copper is

$Q=m C_s \Delta T$

where

m = 0.25 kg is the mass of the sample

C_s = 390 J/kg C is the specific heat of copper

$\Delta T = 25.0^{\circ}C - 75.0^{\circ}C=-50.0^{\circ}$ is the change in temperature of the copper

Substituting the values into the formula, we find:

$Q=(0.25 kg)(390 J/kg C)(-50.0^{\circ}C)=-4875 J$

And the negative sign means the heat has been released by the substance.

4. 2.29 kg

Similarly to the previous part, the amount of heat released from the sample of mercury is

$Q=m C_s \Delta T$

where in this problem we know

m is the mass of the sample

C_s = 140 J/kg C is the specific heat of mercury

$\Delta T = 15.0^{\circ}C - 40.0^{\circ}C=-25.0^{\circ}$ is the change in temperature of the sample

Q = -8,000 J is the heat released by the sample

Solving the formula for m, we find:

$m=\frac{Q}{C_s \Delta T}=\frac{-8000 J}{(140 J/kg C)(-25.0^{\circ})}=2.29 kg$

When the two samples of water (hot water and cold water) are put in contact, thermal energy is transferred from the hot water to the cold water. This occurs because heat always flows from a hotter object to a colder object. The heat is transferred by collision between the molecules: the molecules of the hot water have on average more kinetic energy than the molecules of cold water, so when they collide to each other, the molecules of hot water transfer energy to the molecules of cold water. As a result, the kinetic energy of the molecules of cold water increases, and therefore the temperature of the cold water increases, while the temperature of the hot water decreases. This process lasts until the molecules of the two samples have same average kinetic energy: when this occurs, the two samples have same temperature, so the heat flow stops.

answered: Guest

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

answered: Guest

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