Odpowiedź:
[tex]Zad.1) 40 \frac{J}{kg * C}\\Zad.2) 25 \frac{J}{kg * C}\\Zad.3) 20 \frac{J}{kg * C}[/tex]
Wyjaśnienie:
Zad.1
m = 15 kg (masa)
Q = 5400 J ( Ciepło = Energia )
ΔT = 24°C - 15°C = 9°C ( przyrost temperatury )
[tex]c_{w} = ?[/tex] ( ciepło właściwe )
[tex]Q = c_{w} *m*[/tex] ΔT
5400J = [tex]c_{w}[/tex] * 15 kg * 9°C
5400J = [tex]c_{w}[/tex] * 135 kg* °C /: 135 kg* °C
[tex]c_{w}[/tex] = 40 [tex]\frac{J}{kg * C}[/tex]
Zad.2
m = 10 kg (masa)
Q = 25kJ = 25000 J ( Ciepło = Energia )
ΔT = 130°C - 30°C = 100°C ( przyrost temperatury )
[tex]c_{w} = ?[/tex] ( ciepło właściwe )
[tex]Q = c_{w} *m*[/tex] ΔT
25000J = [tex]c_{w}[/tex] * 10 kg * 100°C
25000J = [tex]c_{w}[/tex] * 1000 kg* °C /: 1000 kg* °C
[tex]c_{w}[/tex] = 25 [tex]\frac{J}{kg * C}[/tex]
Zad.3
m = 3 kg (masa)
Q = 3,6 kJ = 3600 J ( Ciepło = Energia )
ΔT = 90°C - 30°C = 60°C ( przyrost temperatury )
[tex]c_{w} = ?[/tex] ( ciepło właściwe )
[tex]Q = c_{w} *m*[/tex] ΔT
3600J = [tex]c_{w}[/tex] * 3 kg * 60°C
3600J = [tex]c_{w}[/tex] * 180 kg* °C /: 180 kg* °C
[tex]c_{w}[/tex] = 20 [tex]\frac{J}{kg * C}[/tex]
