How much energy does the water to transfer to the room as it cools? The specific heat capacity of water is 4200J/kg/'C. The mass is 500ml of water at a temp of 50'C. The starting temp is 50'C&end temp 20'C. Cannot work this out for the life of me!
-
Energy = mass x specific heat capacity x change in temp
E = 0.5 kg x 4200 x -30
E = -63000 J
The water loses 63000 J of thermal energy to its surroundings.
E = 0.5 kg x 4200 x -30
E = -63000 J
The water loses 63000 J of thermal energy to its surroundings.
-
The formula for this change is Q = m(Ti - Tf)Cp where m = mass, Ti - beginning temp, Tf = final temp, and Cp is the specific heat capacity. Since the density of water is 1, we know that 500mL has a mass of 500-g or 0.5-Kg. Q = 0.5=Kg x (50 oC - 20 oC) x 4200-J/Kg*oC = 63,000-J or 63-KJ
-
mcΔT = Q
0.500kg x 4,200kJ/kg/°C x 30°C ΔT = 63kJ of heat energy transferred.
0.500kg x 4,200kJ/kg/°C x 30°C ΔT = 63kJ of heat energy transferred.
-
temp. fallen=30 C
heat transferred to room=30*4200*1/2(as 500 ml=1/2kg.)
Heat transferred=63000 Joules.........
am i right??
heat transferred to room=30*4200*1/2(as 500 ml=1/2kg.)
Heat transferred=63000 Joules.........
am i right??
-
first 2 right 3rd wrong