V1/T1=V2/T2
V2=(V1)(T2)/T1
Plug in values given (for the temp you can either turn 300K to 27°C or turn 132°C into kelvin
V2= 4400 mL= 4.4L
Answer:
At the start of the process, the volume not occupied by the water is 2 m3
Explanation:
At the start of the process you have a half full tank. It means that also a half is empty (not occupied by water).
Since the volume is 4 m3, 2 m3 are full (occupied by water) and 2 m3 (not occupied by water).
The volume in time will be
![V(t)=V_0+(f_i-f_o)*t\\\\V(t) = 2 +(6.33/1000-3.25/1000)*t=2+0.00308*t \, \, [m3]](https://tex.z-dn.net/?f=V%28t%29%3DV_0%2B%28f_i-f_o%29%2At%5C%5C%5C%5CV%28t%29%20%3D%202%20%2B%286.33%2F1000-3.25%2F1000%29%2At%3D2%2B0.00308%2At%20%5C%2C%20%5C%2C%20%5Bm3%5D)
The first and Third graph
Answer: c = 710 J/kg°C or 0.71 J/g°C
Explanation: Heat is expressed in the formula Q = mc∆T. Derive to find the specific heat c. So the formula will become c = Q / m∆T
c = Q / m∆T
= 42600 J / 2 kg ( 55°C - 25°C )
= 710 J /kg°C
Or can be expressed by converting kg to g.
c = 0.71 J /g°C