Question

A 50 cm^3 block of iron is removed from an 800 degrees Celsius furnance and immediately dropped into 200 mL of 20 degrees Celsius water. What percentage of the water boils away?

Answer 1

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

$m_1\times C_1\times \Delta T = M\times C_w\times \Delta T + m_2\times L_v$

$393.7\times 0.45\times (800-100) = 200\times 4.18\times(100-20) + m_2\times 2260$

m₂ = 25.28 g

25.28 water will be vaporized

% of water boils away =$\dfrac{25.28}{200}\times 100$

 % of water boils away= 12.64 %