1. [tex]2.3\cdot 10^5 J[/tex]
The amount of heat needed to melt a substance at its melting point is:
[tex]Q=m\lambda_f[/tex]
where
m is the mass of the substance
[tex]\lambda_f[/tex] is the latent heat of fusion of the substance
For the sample of lead in this problem,
m = 10.0 kg
[tex]\lambda_f = 23000 J/kg[/tex] (latent heat of fusion of lead)
Substituting,
[tex]Q=(10.0 kg)(23000 J/kg)=2.3\cdot 10^5 J[/tex]
2. [tex]2.96\cdot 10^6 J[/tex]
The amount of heat needed to evaporate a substance at its boiling point is:
[tex]Q=m\lambda_v[/tex]
where
m is the mass of the substance
[tex]\lambda_v[/tex] is the latent heat of vaporization of the substance
For the sample of mercury in this problem,
m = 10.0 kg
[tex]\lambda_v = 2.96\cdot 10^5 J/kg[/tex] (latent heat of vaporization of mercury)
Substituting,
[tex]Q=(10.0 kg)(2.96\cdot 10^5 J/kg)=2.96\cdot 10^6 J[/tex]
3. [tex]-4875 J[/tex]
The amount of heat released from the sample of copper is
[tex]Q=m C_s \Delta T[/tex]
where
m = 0.25 kg is the mass of the sample
[tex]C_s = 390 J/kg C[/tex] is the specific heat of copper
[tex]\Delta T = 25.0^{\circ}C - 75.0^{\circ}C=-50.0^{\circ}[/tex] is the change in temperature of the copper
Substituting the values into the formula, we find:
[tex]Q=(0.25 kg)(390 J/kg C)(-50.0^{\circ}C)=-4875 J[/tex]
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
[tex]Q=m C_s \Delta T[/tex]
where in this problem we know
m is the mass of the sample
[tex]C_s = 140 J/kg C[/tex] is the specific heat of mercury
[tex]\Delta T = 15.0^{\circ}C - 40.0^{\circ}C=-25.0^{\circ}[/tex] 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:
[tex]m=\frac{Q}{C_s \Delta T}=\frac{-8000 J}{(140 J/kg C)(-25.0^{\circ})}=2.29 kg[/tex]
5.
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.
Answer:
1.Heat needed to melt [tex]10.0[/tex]kg of lead is [tex]$2.3 \cdot 10^{5}$[/tex]J.
2. Heat needed to vaporize [tex]10.0[/tex]kg of mercury is [tex]$2.96 \cdot 10^{6}$[/tex]J.
3.The heat which flows out of the copper sample is [tex]$-4875[/tex]J.
4.The behavior of the particles before and after thermal equilibrium reached is as the temperature of the hot and cold water remains constant for some time.
Explanation:
1. To find the heat needed to melt lead at its melting point is given by a formula,
[tex]$Q=m \lambda_{f}$[/tex]
As,
[tex]\lambda_{f}$=[/tex]Latent heat of the substance
[tex]m=[/tex]Mass of the substance
Given,
mass of lead[tex]=10.0[/tex]kg
We know that,the latent heat of lead is [tex]$23000 $[/tex] J/kg
So apply in formula,
[tex]$Q=(10.0 k g)(23000 J / k g)$[/tex]
[tex]=$2.3 \cdot 10^{5}$J[/tex].
2. To find, heat needed to vaporize mercury at its boiling point is given by a formula,
[tex]$Q=m \lambda_{v}$[/tex]
As,
[tex]$ \lambda_{v}=$[/tex]Latent heat of the substance.
[tex]m=[/tex]mass of the substance.
Mass of mercury[tex]=10.0[/tex]kg
We know that,the latent heat of Mercury is [tex]$2.96 \cdot 10^{5}$[/tex]J/kg
Apply in formula,
[tex]$Q=(10.0 \mathrm{~kg})\left(2.96 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\right)[/tex]
[tex]=2.96 \cdot 10^{6} $[/tex]J.
3. The amount of heat released from the sample of copper given by,
[tex]$Q=m C_{s} \Delta T$[/tex]
In this,
[tex]m=[/tex]Mass of the copper sample
The specific heat of mercury,
[tex]$C_{s}=140 \mathrm{~J} / \mathrm{kgC}$[/tex]
Change in temperature of copper sample,
[tex]$\Delta T=15.0^{\circ} \mathrm{C}-40.0^{\circ} \mathrm{C}[/tex]
[tex]=-25.0^{\circ}$[/tex]
Also we know that,
[tex]$Q=-8,000 \mathrm{~J}$[/tex]
Mass value is as,
[tex]$m=\frac{Q}{C_{s} \Delta T}$[/tex]
Apply the value in formula we get,
[tex]$=\frac{-8000 . J}{(140 J / \mathrm{kgC})\left(-25.0^{\circ}\right)}[/tex]
[tex]=2.29 \mathrm{~kg}$[/tex].
4. The behavior of the particles before and after thermal equilibrium,
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