57)
m(water) = 254 g
T(water) = 39.0 oC
C(water) = 4.184 J/goC
m(Cu) = 43.9 g
T(Cu) = 135 oC
C(Cu) = 0.385 J/goC
T = to be calculated
Let the final temperature be T oC
use:
heat lost by Cu = heat gained by water
m(Cu)*C(Cu)*(T(Cu)-T) = m(water)*C(water)*(T-T(water))
43.9*0.385*(135.0-T) = 254.0*4.184*(T-39.0)
16.9015*(135.0-T) = 1062.736*(T-39.0)
2281.7025 - 16.9015*T = 1062.736*T - 41446.704
T= 40.5 oC
Answer: B
Only 1 question at a time please
57) A 43.9-8 piece of copper (CCu= 0.385 J/gºC) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that...
would you do all the questions please
56) How much heat energy is required to raise the temperature of 0.298 mole of water from 35.93°C to 79.84" C? The specific heat capacity of water = 4.18 J/g"C A) 13.1 B) 54.7 J C) 985 J D) 2105 J E) 3310J 57) A 43.9-g piece of copper (Ccu= 0.385 J/g*C) at 135.0 C is plunged into 254 g ofwater at 39.0°C. Assuming that no heat is lost to the surroundings, what...