Answer:
the final temperature is T f = 64.977 ° C≈ 65°C
Step-by-step explanation:
Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:
Q coffee + Q ice = Q surroundings =0 (insulated)
We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).
The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)
The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)
therefore
m co * c co * (T fco - T ico) + m ice * L + m ice * c wat * (T fwa - T iwa) = 0
assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water
d co = dw = 1 gr/cm³
therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr
m co * c wat * (T f - T ico) + m ice * L + m ice * c wat * (T f - T iwa) = 0
m co * c wat * T f+ m ice * c wat * T f = m ice * c wat * T iwa + m co * c wat * Tico -m ice * L
T f = (m ice * c wat * T iwa + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )
replacing values
T f = (11 g * 4.186 J/g°C * 0°C + 106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C + 106 g * 4.186 J/g°C* ) = 64,977 ° C
T f = 64.977 ° C