# Question 1
For Air
ΔT = 10
floor area = 15m^2
height = 2.5m
heater rated = 1668W
density = 1.25kg/m^3
specific heat = 1kj/kg C

Volume = area * height = 15 * 2.5 = 37.5m^3

mass = density * volume = 1.25 * 37.5 = 46.875 kg
room specific heat = specific heat * mass = 1 * 46.875 = 46.875kj
ΔQ for ΔT = ΔT * room specific heat = 10 * 46.875 = 468.75kj = 468 750j

time for ΔQ = Q / heater rated = 468 750 / 1668 = ~281s = 4.6 min

The room full of Air would take 4.6 min. The room full of water (assuming a density of 997kg/m^3 and specific heat of 4 184j/kg C) would take about 15630 min.

For Water
ΔT = 10
Volume = 37.5m^3
heater rated = 1668W
density = 997 kg/m^3
specific heat = 4 184 j/kg C

mass = Volume * density = 37.5 * 997 = 37 387.5kg
room specific heat = specific heat * mass = 4 184 * 37 387.5 = 15 6429 300 j
ΔQ for ΔT = ΔT * room specific heat = 10 * 15 6429 300 = 1 564 293 000 j

time for ΔQ = Q / heater rated = 1 564 293 000 / 1668 = 937825.53s = 15 630m

# Question 2
area = 137m^2
heat demand = 150kWh/m^2
cost of switch = 119 460 sek
electricity price = 1540 sek/MWh
Ideal efficency
Cold temp = 0C = 273
Hot temp = 60C = 333

House heat demands annual = area * heat demand = 137 * 150 = 20 550 kWh = 20.550 MWh
cost before switching = House heat demands annual * electricity price = 20.550 * 1540 = 31 647 sek

$$
efficiency = {T_h \over ΔT} = {333 \over 60} = 5.55
$$
heat pump W = House heat demands annual / efficiency = 20.550 / 5.55 = 3.7027 MWh
cost after switching = heat pump W * electricity price = 3.7027 * 1540 = 5 702 sek
Per year saving = cost before switching - cost after switching = 31 647 - 5 702 = 25 945 sek
years before payed off = cost of switch / Per year saving = 119 460 / 25 945 = 4.6 years

# Question 3
On consumption = 109 W = 109Wh
Off consumption = 0 W
8 hours on
electricity cost = 2621 sek/MWh

day usage = 8 * 109Wh = 872 Wh
year = 365 * 872 = 318280 Wh = 0.31828 MWh
year cost = 2621 * 0.31 = 834.211 sek

# Question 4
$$
f(x) = \left(7.93 - 8.67cos\left({2π \over 365} \cdot (x-25.2)\right)\right)
$$
the expression dose not go above 20. So we can just subtract the result form 20 to get the temperature difference between indoors and out doors
$$
G(x) = 20-\left(7.93 - 8.67cos\left({2π \over 365} \cdot (x-25.2)\right)\right)
$$
For every degree of difference between the indoor and outdoor temp 1W/m^3 is needed for heating.

To get the energy needed over the year we intergrade the function.

$$
\int_1^{365} G(x)dx = 4405 W = 4.385 kW
$$
So the energy needed by a home is 4.385kW/m^2

# Question 5
Hot temp = 151
Cold temp = 12

$$
ε_{max} = {ΔT \over T_h} = {139 \over 424} = 0.3278 = 32.7 \%
$$
If Cold temp = 80

$$
ε_{max} = {ΔT \over T_h} = {71 \over 424} = 0.1674 = 16.7 \%
$$
efficiency would decrease to 16.7 %