Answer:
Q = 4019.4 J
Explanation:
Given data:
Mass of ice = 20.0 g
Initial temperature = -10°C
Final temperature = 89.0°C
Amount of heat required = ?
Solution:
specific heat capacity of ice is 2.03 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 89.0°C - (-10°C)
ΔT = 99°C
Q = 20.0 g ×2.03 J/g.°C × 99°C
Q = 4019.4 J
Answer:
ΔT = Tfinal − Tinitial = 150°C − 35.0°C = 125°C
given the specific heat of iron as 0.108 cal/g·°C
heat=(100.0 g)(0.108 cal /g· °C )(125°C) =
100x 0.108x125= 1350 cal
Explanation:
A) particles are close together in random positions with about equal kinetic energy and intermolecular forces.
These points are about liquid state.
B) particles are close together in fixed positions with low kinetic energy
These points satisfy the qualities of Solid state
C)particles are far apart with greater kinetic energy and low intermolecular forces.
The above qualities are for Gaseous state of matter
A) Liquid
B)Solid
C)Gas
Combustion reaction
Key: O2
O2 is normally in a chemical formula when you are used to burn anything, so basically, anything with O2 involves burning.
