# 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?

Sep 16, 2016

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

$T = {42}^{o} C$

So the temperature change:

ΔT=42^oC

#### Explanation:

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

${Q}_{t o t} = {Q}_{1} + {Q}_{2}$

Initial heat (before mixing)

Where ${Q}_{1}$ is the heat of water and ${Q}_{2}$ the heat of the object. Therefore:

${Q}_{1} + {Q}_{2} = {m}_{1} \cdot {c}_{{p}_{1}} \cdot {T}_{1} + {m}_{2} \cdot {c}_{{p}_{2}} \cdot {T}_{2}$

Now we have to agree that:

• The heat capacity of water is:

${c}_{{p}_{1}} = 1 \frac{k c a l}{k g \cdot K} = 4 , 18 \frac{k J}{k g \cdot K}$

• The density of water is:

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

So we have:

${Q}_{1} + {Q}_{2} =$

$= 37 k g \cdot 4 , 18 \frac{k J}{k g \cdot K} \cdot \left(0 + 273\right) K + 2 k g \cdot 12 \frac{k J}{k g \cdot K} \cdot \left(315 + 273\right) K$

${Q}_{1} + {Q}_{2} = 56334 , 18 k J$

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:

${Q}_{1} + {Q}_{2} = {m}_{1} \cdot {c}_{{p}_{1}} \cdot T + {m}_{2} \cdot {c}_{{p}_{2}} \cdot T$

Use equation to find final temperature:

${Q}_{1} + {Q}_{2} = T \cdot \left({m}_{1} \cdot {c}_{{p}_{1}} + {m}_{2} \cdot {c}_{{p}_{2}}\right)$

$T = \frac{{Q}_{1} + {Q}_{2}}{{m}_{1} \cdot {c}_{{p}_{1}} + {m}_{2} \cdot {c}_{{p}_{2}}}$

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

$T = {315}^{o} K$

$T = 315 - 273 = {42}^{o} C$

Provided the pressure is atmospheric, the water did not evaporate, since its boiling point is ${100}^{o} C$. The final temperature is:

$T = {42}^{o} C$

So the temperature change:

ΔT=|T_2-T_1|=|42-0|=42^oC