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* c_1* (T_f-T_1)=-m_2* c_2* (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* 1.87J/g^oC* (30-100)^oC=-200000g* 2.06J/g^oC* (30-0)^oC`

`m_1=94423.224g=94.42kg`

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