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
Answer:
The specific heat of iron is 0.45 J/g.°C
Explanation:
The amount of heat absorbed by the metal is given by:
heat = m x Sh x ΔT
From the data, we have:
heat = 180.8 J
mass = m = 22.44 g
ΔT = Final temperature - Initial temperature = 39.0°C - 21.1 °C = 17.9°C
Thus, we calculate the specific heat of iron (Sh) as follows:
Sh = heat/(m x ΔT) = (180.8 J)/(22.44 g x 17.9°C) = 0.45 J/g.°C
Answer: 3.35x10²³atoms H2
Explanation: solution attached:
Convert mass of Al to moles
Do the mole to mole ratio between Al and H2
Convert moles of H2 to atoms using Avogadro's number.
