Answer:
7.2g
Explanation:
From the expression of latent heat of steam, we have
Heat supplied by steam = Heat gain water + Heat gain by calorimeter
mathematically,
[tex]m_{s}c_{w} \alpha _{w} [/tex] + [tex]m_{s}l[/tex]=[tex]m_{w}c_{w} \alpha _{w}[/tex] +[tex]m_{c}c_{c} \alpha c_{c} [/tex]
L=specific latent heat of water(steam)=2268J/g
[tex]c_{w}[/tex]=specific heat capacity=4.2J/gK
[tex]c_{c}[/tex]=specific heat capacity of calorimeter =0.9J/gk
[tex]m_{w}[/tex]=280g
[tex]m_{c}[/tex]=38g
α=change in temperature
[tex]\alpha _{c}[/tex]=(40-25)=15
[tex]\alpha _{w}[/tex]=(40-25)=15
[tex]\alpha _{s}[/tex]=(100-40)=60
Note: the temperature of the calorimeter is the temperature of it content.
From the equation, we can make [tex]m_{s}[/tex] the subject of formula
[tex]m_{s}=\frac{m_{w}c_{w} \alpha +m_{c}c_{c}\alpha}{c_{w}\alpha +l}[/tex]
Hence
[tex]m_{s}=\frac{(280*4.2*15) +(38*0.9*15)}{(4.2*60) +2268} \\m_{s}=\frac{18153}{2520}\\ m_{s}=7.2g[/tex]
Hence the amount of steam needed is 7.2g
The required amount of steam for the system to reach a final temperature of 40.0°C is 7.2g.
According to the conservation of energy:
Heat lost by the steam = heat gained by water + heat gained by aluminum cup
The steam condenses to water at 100°C, so latent heat energy will also be involved.
Let m₁, m₂ and m₃ be the mass of steam , aluminum cup and water.
and c₁, and c₂ be the repsective specific heats of water and aluminum.
L be the latent heat of evaporaion of water.
Heat lost by steam is:
Q = m₁c₁ΔT + m₁L
Q = m₁×4.2×(100-40) + m₁×2268
Q = 2520m₁
heat gained by the cup and water:
Q = m₂c₁ΔT + m₃c₃ΔT
Q = 38×0.9×(40-25) + 280×4.2×(40-25)
Q = 18153J
now,
2520m₁ = 18153
m₁ = 7.2g
So, 7.2g steam is needed.
Leatn more about latent heat:
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