What mass of water can be heated from 25 degrees C to 75.0 degrees C by the addition of 7.96 x 10 4 Joules? About the author Kennedy
Explanation: Specific heat of water is = 1cal/°C = 4.18J/K That means water requires 4.18 Joules of heat energy to raise the temperature of 1 g of water by 1°C or 1K. Similarly water will lose 4.18 Joules of heat energy for every gram of water by 1°C. Therefore (625 x 4.18) Joules of heat energy is lost for every drop in 1°C of water. So number degrees it gets cooled for 7.96 x 10⁴ Joules = 79600/(625 x 4.18) =30.47°C So the final temperature of 625 g of water = (75–30.47)= 44.53°C c=4200 m=0.625 t1=75 dQ=7,96*10^4=79600J dt=t1-t2 d-delta(change) Q=cm(dt) dt=Q/cm=79600/(4200*0.625)=30.3238 t2=75-30,3=44,7°C Reply
Answer:
27.1 g is the correct answer
Explanation:
Specific heat of water is = 1cal/°C = 4.18J/K
That means water requires 4.18 Joules of heat energy to raise the temperature of 1 g of water by 1°C or 1K.
Similarly water will lose 4.18 Joules of heat energy for every gram of water by 1°C.
Therefore (625 x 4.18) Joules of heat energy is lost for every drop in 1°C of water.
So number degrees it gets cooled for 7.96 x 10⁴ Joules
= 79600/(625 x 4.18)
=30.47°C
So the final temperature of 625 g of water = (75–30.47)= 44.53°C
c=4200 m=0.625 t1=75 dQ=7,96*10^4=79600J
dt=t1-t2 d-delta(change) Q=cm(dt)
dt=Q/cm=79600/(4200*0.625)=30.3238
t2=75-30,3=44,7°C