Apot on the stove contains 200 g of water at 20°c. an unknown mass of ice that is originally at −10°c is placed in an identical pot on the stove. heat is then added to the two samples of water at precisely the same constant rate. assume that this heat is transferred immediately to the ice or water (in other words, neglect the increase in temperature for the pot). we will also neglect evaporation. the ice melts and becomes water, and you observe that both samples of water reach 55.0°c at the same time.

(a) how does the mass of the original piece of ice in the second pot compare to the mass of the water in the first pot?

the mass of the ice is smaller than the mass of the water.

the mass of the ice is larger than the mass of the water

the mass of the ice is the same as the mass of the water.

there's not enough information to answer this question.

(b) which system will reach 80°c first?

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

so the time that takes to reach 80°C is higher for water , thus the ice mass reaches it first.

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )