Answer:
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.
