# An object with a mass of 1 kg, temperature of 155 ^oC, and a specific heat of 32 (KJ)/(kg*K) is dropped into a container with 15 L  of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

Feb 10, 2017

At the end the water is at 52.32 °C

#### Explanation:

The object temperature decreases from 155°C to the temperature T1 giving heat to the water
 Q= c_(po) M_0(T_0-T_1)= (32kJ)/(kgK)1kg (155-T_1)°C.
Water takes this heat, warming from t= 0°C to the temperature T1.
 Q= c_(pw)M_w(T_1-0°C)=(4.186 kJ)/(kgK) 15kg ((T_1-0)°C
(as one liter of water is one kg of water) equaling the lost heat with the gained heat, and resolving as a function of ${T}_{1}$, you find T_1= 52.32 °C

as $32 \left(155 - {T}_{1}\right) = 4.186 \times 15 \times \left({T}_{1} - 0\right)$

or $4.186 \times 15 \times {T}_{1} + 32 {T}_{1} = 32 \times 155$

or $\left(62.79 + 32\right) {T}_{1} = 94.79 {T}_{1} = 4960$

or T_1=52.32 °C

Warning: There doesn't exist a solid object with a specific heat of $\frac{32 k J}{k g K}$ . Only the ammonia among the liquid substances has specific heat bigger than water, which is one