Question

mix some 40 degree water with some 30 degree water and youll get water at?
a) 33.3 C
b)35 C
c) 37 C
d) none of these

Answer 1

From what I remember in thermal\[mC _{ice} +mC_{ice}(T _{f}-T_{i})+mC_{water}(T_{f}-T_{i}) =0\]C relates to the specific heat of ice and water respectively, Tf is the final temperature and Ti are the initial temperatures of the ice and water respectively. This is an equation that says in an isolated system there can be no change in the total heat of the system (Q).

C(ice) = 2.108 [kJ]/[kg][K]
C(water) = 4.187 [kJ]/[kg][K]
Ti(ice) = 273 + 30 (conversion to Kelvin degrees) = 303 K
Ti(water) = 273 + 40 = 313 K
m = 1 kg (this is an assumption I've made since it's not mentioned)

Plug and chug with these numbers and you should arrive at 37 C. This is assuming I've remembered my schooling correctly! Double-checking never hurts. Intuitively this makes sense because it takes more energy to change the temperature of water than ice. So if the masses are equal the final result should be closer to the initial temp of the water. Hope this helps!

Answer 2

It depends on mesessä of water. If amounts are etualalla, temperature is 35.
But any amounts are not mentioned here.