Answer:
98°C
Explanation:
Given data:
Volume of water = 20.0 mL
Temperature = 32°C
Mass of skillet = 1.15 kg
Change in temperature = ?
Solution:
Q = m.c.ΔT
ΔT = 100°C - 32°C
ΔT = 68°C
Q = m.c.ΔT
Q = 20 g× 4.184 j/g.°C× 68°C
Q = 5690.24 j
Heat of vaporization of water = 2257 j/g
q = 2257 j/g × 20 g
q = 45140 j
Total heat gained by water= 5690.24 j + 45140 j
Total heat gained by water= 50830.24 j
Number of moles of iron:
Number of moles = mass/ molar mass
Number of moles = 1150/55.85
Number of moles = 20.59 mol
Molar heat capacity of iron is =25.19 j/mol.°C
50830.24 j/ 25.19 j/mol.°C/20.59 mol
98°C
