Answer:
7.19 * 10^14J
Explanation:
Given that
Density of water Pwater= 1000kg/m3
R=2.1km = 2.1*10^3m
H= 2.3cm. = 2.3*10^-2m
Lv water= 2256 * 10^3J/kg
First, mass of water need to be calculated, using an imaginary cylinder
Density= Mass /Volume
Mass= Density* Volume
Volume of a cylinder= πR2h
Therefore mass= PπR2H
= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)
= 3.18 *10^8
Heat Released Qv = mLV
= 3.18*10^8 * 2236*10^3
= 7.19 * 10^14J
It has three significant figure
Answer:
A) T1 = 269.63 K
T2 = 192.59 K
B) W = -320 KJ
Explanation:
We are given;
Initial volume: V1 = 7 m³
Final Volume; V2 = 5 m³
Constant Pressure; P = 160 KPa
Mass; m = 2 kg
To find the initial and final temperatures, we will use the ideal gas formula;
T = PV/mR
Where R is gas constant of helium = R = 2.0769 kPa.m/kg
Thus;
Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K
Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K
B) world one is given by the formula;
W = P(V2 - V1)
W = 160(5 - 7)
W = -320 KJ
