To convert 78.1 g of water at 0° C to Ice at -57.1°C; we can do it in steps;
1. Water at 0°C to ice at 0°C
The heat of fusion of ice is 334 J/g;
Heat = 78.1 × 334 = 26085.4 Joules
2. Ice at 0°C to -57.1°C
Specific heat of ice is 2.108 J/g
Heat = 78.1 × 2.108 J/g = 164.6348 Joules
Thus the total heat energy released will be; 26085.4 + 164.6348
= 26250.0348 J or 26.250 kJ
If you want an example then a car passing by a standing man is the best one !!
Answer:
C
Explanation:
acids are corrosive since they tend to destroy every single thing they do get a big example is the acidic rain which tends to corrode iron sheet thus making them to appear worn out and full of rust
Hello friend ☺
ΔH = MCΔT
ΔH = to the amount of energy or change in energy (J)
mass of water
C = waters specific heat capacity
ΔT = change in temperature
and so ΔH = 25 × 4.18 × ( 112-67 ) J = 4702.5 J
Thanks ❤
