1 answer

An object with a mass of #6 kg#, temperature of #173 ^oC#, and a specific heat of #14 J/(kg*K)# is

Question:

An object with a mass of #6 kg#, temperature of #173 ^oC#, and a specific heat of #14 J/(kg*K)# is dropped into a container with #32 L # of water at #0^oC #. Does the water evaporate? If not, by how much does the water's temperature change?

Answers

The water does not evaporate and the change in temperature is #=0.11ªC#

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=173-T#

# m_o C_o (DeltaT_o) = m_w C_w (DeltaT_w)#

#C_w=4.186kJkg^-1K^-1#

#C_o=0.014kJkg^-1K^-1#

#6*0.014*(173-T)=32*4.186*T#

#173-T=(32*4.186)/(6*0.014)*T#

#173-T=1594.7T#

#1595.7T=173#

#T=173/1595.7=0.11ºC#

As #T<100ºC#, the water does not evaporate.

.

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