Question

An object with a mass of `2 kg`, temperature of `315 ^oC`, and a specific heat of `12 (KJ)/(kg*K)` is dropped into a container with `37 L` of water at `0^oC`. Does the water evaporate? If not, by how much does the water's temperature change?

Answer 1

The water does not evaporate. The final temperature of the water is:

`T=42^oC`

So the temperature change:

`ΔT=42^oC`

Explanation:

The total heat, if both remain in the same phase, is:

`Q_(t ot)=Q_1+Q_2`

Initial heat (before mixing)

Where `Q_1` is the heat of water and `Q_2` the heat of the object. Therefore:

`Q_1+Q_2=m_1*c_(p_1)*T_1+m_2*c_(p_2)*T_2`

Now we have to agree that:

• The heat capacity of water is:

`c_(p_1)=1 (kcal)/(kg*K)=4,18(kJ)/(kg*K)`

• The density of water is:

`ρ=1 (kg)/(lit)=>1lit=1kg->` so kg and liters are equal in water.

So we have:

`Q_1+Q_2=`

`=37kg*4,18(kJ)/(kg*K)*(0+273)K+2kg*12(kJ)/(kg*K)*(315+273)K`

`Q_1+Q_2=56334,18kJ`

Final heat (after mixing)

• The final temperature of both the water and the object is common.

`T_1'=T_2'=T`

• Also, the total heat is equal.

`Q_1'+Q_2'=Q_1+Q+2`

Therefore:

`Q_1+Q_2=m_1*c_(p_1)*T+m_2*c_(p_2)*T`

Use equation to find final temperature:

`Q_1+Q_2=T*(m_1*c_(p_1)+m_2*c_(p_2))`

`T=(Q_1+Q_2)/(m_1*c_(p_1)+m_2*c_(p_2))`

`T=(56334,18)/(37*4,18+2*12)(kJ)/(kg*(kJ)/(kg*K)`

`T=315^oK`

`T=315-273=42^oC`

Provided the pressure is atmospheric, the water did not evaporate, since its boiling point is `100^oC`. The final temperature is:

`T=42^oC`

So the temperature change:

`ΔT=|T_2-T_1|=|42-0|=42^oC`