Question

In a container of negligible mass, 020 kg of ice at an initial temperature of - 40.0 oC is mixed with a mass m of water that has an initial temperature of 80.0 oC. No heat is lost to the surroundings. If the final temperature of the system is 20.0 oC, what is the mass m of the water that was initially at 80.0 oC

Answer 1

The mass is
`m_w = 0.599 \ kg`

Step-by-step explanation:

From the question we are told that

The mass of ice is
`m_c = 0.20 \ kg`

The initial temperature of the ice is
`T_i = -40.0 ^oC`

The initial temperature of the water is
`T_(iw) = 80^o C`

The final temperature of the system is
`T_f = 20^oC`

Generally according to the law of energy conservation,

The total heat loss is = total heat gained

Now the total heat gain is mathematically represented as

`H = H_1 + H_2 + H_3`

Here
`H_1` is the energy required to move the ice from
`-40^oC \to 0^oC`

And it mathematically evaluated as

`H_1 = m_c * c_c * \Delta T`

Here the specific heat of ice is
`c_c = 2100 \ J \cdot kg^(-1) \cdot ^oC^(-1)`

So

`H_1 = 0.20 * 2100 * (0-(-40))`

`H_1 = 16800\ J`

`H_2` is the energy to melt the ice

And it mathematically evaluated as

`H_2 = m * H_L`

The latent heat of fusion of ice is
`H_L = 334 J/g = 334 *10^(3) J /kg`

So

`H_2 = 0.20 * 334 *10^(3)`

`H_2 = 66800 \ J`

`H_3` is the energy to raise the melted ice to
`20^oC`

And it mathematically evaluated as

`H_3 = m_c * c_w * \Delta T`

Here the specific heat of water is
`c_w= 4190\ J \cdot kg^(-1) \cdot ^oC^(-1)`

`H_3 = 0.20 * 4190* (20-0))`

`H_3 = 16744 \ J`

So

`H = 16800 + 66800 + 16744`

`H = 100344\ J`

The heat loss is mathematically evaluated as

`H_d = m * c_h ( 80 - 20 )`

`H_d = m_w * 4190 * ( 80 - 20 )`

`H_d = 167600 m_w`

So

`167600 m_w = 100344`

=>
`m_w = 0.599 \ kg`