Question about heat/energy?
"A 100.0 gram sample of water at 27.0 degrees C is poured into a 70.1 gram sample of water at 89.0 degrees C. What will be the final temperature of the water?"
Please explain how you got the answer too.
"A 100.0 gram sample of water at 27.0 degrees C is poured into a 70.1 gram sample of water at 89.0 degrees C. What will be the final temperature of the water?"
Please explain how you got the answer too.
Gulizar Ozveri
Favorite Answer
since both of the samples are water,
we find the energies: Q=m*c*T
100*27*1cal/g*C = 2700 cal
70.1*89*1cal/g*C= 6239 cal
add them, 8939 cal and then divide by the total mass, 8939 cal/ 170.1 g = 52.55 cal/g
in order to find the temperature we have to divide by the specific heat capacity 1 cal/g*C
we get 52.55 degrees C.