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?

1 Answer
Andrea B. · Shwetank Mauria
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(155-T_1)=4.186xx15xx(T_1-0)

or 4.186xx15xxT_1+32T_1=32xx155

or (62.79+32)T_1=94.79T_1=4960

or T_1=52.32 °C

Warning: There doesn't exist a solid object with a specific heat of (32kJ)/(kgK) . Only the ammonia among the liquid substances has specific heat bigger than water, which is one

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