Question

Steam (water vapor) at 100 degrees Celsius is added to a thermally insulated container with 200 kg of ice at zero degrees Celsius. The final mixture is water at 30 degrees Celsius. What was the initial mass of the steam?

Answer 1

Answer : The initial mass of the steam is 94.42 kg.

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of water vapor = $1.87J/g^oC$

$c_2$ = specific heat of solid water = $2.06J/g^oC$

$m_1$ = mass of steam = ?

$m_2$ = mass of solid water = 200 kg = 200000 g

$T_f$ = final temperature of water = $30^oC$

$T_1$ = initial temperature of water vapor = $100^oC$

$T_2$ = initial temperature of solid water = $0^oC$

Now put all the given values in the above formula, we get

$m_1\times 1.87J/g^oC\times (30-100)^oC=-200000g\times 2.06J/g^oC\times (30-0)^oC$

$m_1=94423.224g=94.42kg$

Therefore, the initial mass of the steam is 94.42 kg.