An object with a mass of 16 kg, temperature of 180 ^oC, and a specific heat of 5 J/(kg*K) is dropped into a container with 48 L of water at 0^oC . Does the water evaporate? If not, by how much does the water's temperature change?

1 Answer
Narad T.
Mar 21, 2018

The water does not evaporate and the change in the tempertaure is =0.07^@C

Explanation:

The heat is transferred from the hot object to the cold water.

Let T= final temperature of the object and the water

For the cold water, Delta T_w=T-0=T

For the object DeltaT_o=180-T

m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)

The specific heat of the object is C_o=0.005kJkg^-1K^-1

The mass of the object is m_0=16kg

The volume of water is V=48L

The density of water is rho=1kgL^-1

The mass of the water is m_w=rhoV=48kg

16*0.005*(180-T)=48*4.186*T

180-T=(48*4.186)/(16*0.005)*T

180-T=2511.6T

2512.6=180

T=180/2512.6=0.07^@C

As the final temperature is T<100^@C, the water will not evaporate. We expect this result as the temperature of the object is not very high and the specific heat is low.

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