# An object with a mass of 4 kg, temperature of 240 ^oC, and a specific heat of 25 J/(kg*K) is dropped into a container with 36 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Jun 19, 2017

water's temperature change = 0,16°C

#### Explanation:

The object decrease its temperature toward a final temperature Tf giving an amount of heat that is:
$Q = m c p \left(T i - T f\right) = 4 k g 25 \frac{J}{k g K} \left(T i - T f\right) = 100 \frac{J}{K} \left(T i - T f\right)$

The water increases its temperature acquiring an egual amount of heat: Q= m cp (Tf-T°)= 36 kg 4186 J/(kg K) (Tf-T°)= 150 000(J)/K(Tf-T°)
Equalizing the two quantities of heat you have:
100 (J)/K(240°C-Tf) =150 000(J)/K(Tf-0°C)
Simplifying
2(240°C-Tf) =3000(Tf-0°C)
and resolving
480 - 2Tf = 3000 Tf
3002 Tf = 480
Tf = 0,16 °C
note that: 36 L of water are 36 kg; specific heat of water is (4186 J)/kg K); the specific heat of the object is really very small