Answer:
Specific heat of water = 33.89 KJ
Explanation:
Given:
mass of water = 81 gram
Initial temperature = 0°C
Final temperature = 100°C
Specific heat of water = 4.184
Find:
Required heat Q
Computation:
Q = Mass x Specific heat of water x (Final temperature - Initial temperature)
Q = (81)(4.184)(100-0)
Q = 33,890.4
Specific heat of water = 33.89 KJ
Answer:
Has size and magnitude whereas a scalar quantity has only size
Saturated hydrocarbons are organic compounds that contain only single bonds between the carbon atoms. They are known to be the simplest organic compounds. They are termed as such because they are saturated with water. Examples are the alkanes (ethane, methane, propane, butane, etc.).
Answer:
5)0.53liters
6)1.634.6°c
Explanation:
5) v1=25liters, t1=12000°c. v2=?, t2=250°c
v2=(250×25)/12000
v2=6250/12000
v2=0.52liters
6)v1=130,v2=85, t1=2500, t2=?
t2=(85×2500)/130
t2=212500/130
t2= 1634.6°c
