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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