Respuesta :
Answer:
% of water boils away= 12.64 %
Explanation:
given,
volume of block = 50 cm³ removed from temperature of furnace = 800°C
mass of water = 200 mL = 200 g
temperature of water = 20° C
the density of iron = 7.874 g/cm³ ,
so the mass of iron(m₁) = density × volume = 7.874 × 50 g = 393.7 g
the specific heat of iron C₁ = 0.450 J/g⁰C
the specific heat of water Cw= 4.18 J/g⁰C
latent heat of vaporization of water is L_v = 2260 k J/kg = 2260 J/g
loss of heat from iron is equal to the gain of heat for the water
[tex]m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v[/tex]
[tex]393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260[/tex]
m₂ = 25.28 g
25.28 water will be vaporized
% of water boils away =[tex]\dfrac{25.28}{200}\times 100[/tex]
% of water boils away= 12.64 %
The percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.
What is heat transfer?
The heat transfer is the transfer of thermal energy due to the temperature difference.The heat flows from the higher temperature to the lower temperature.
The heat transfer of a closed system is the addition of change in internal energy and the total amount of work done by it.
As the initial volume of the iron block is 50 cm³ and the density of the iron is 7.874 g/cm³. Thus the mass of the iron block is,
[tex]m=50\times7.874\\m=393.7\rm g[/tex]
The temperature of the furnace is 800 degrees Celsius and the specific heat of the iron block is 0.45 J/g-C.
As the boiling point of the water is 100 degree Celsius. Thus the heat loss by the block of iron is,
[tex]Q_L=393.7\times0.45\times(800-100)\\Q_L=124015.5[/tex]
The latent heat of the water is 2260 J/g. Thus the heat gain by vaporized water is,
[tex]Q_v=2260\times m_v\\[/tex]
Now the heat gain by the water is equal to the heat loss by the iron block.
As the specific heat of the water is 4.18 J/g-C and the temperature of the water is 20 degrees and volume of water is 200 ml.
Thus heat gain by water can be given as,
[tex]Q_G=Q_L=200\times4.18(100-20)+2260m_v\\124015.5=200\times4.18(100-20)+2260m_w\\m_v=25.28\rm g[/tex]
Thus the total amount of the water boils away is 25.28 grams.
The percentage of the water boils away is,
[tex]p=\dfrac{25.25}{200}\times100\\p=12.64[/tex]
Thus the percentage of the water boils away when the iron block is placed into the water after furnace is 1264%.
Learn more about the heat transfer here;
https://brainly.com/question/12072129
