Given:
Ma = 31.1 g, the mass of gold
Ta = 69.3 °C, the initial temperature of gold
Mw = 64.2 g, the mass of water
Tw = 27.8 °C, the initial temperature of water
Because the container is insulated, no heat is lost to the surroundings.
Let T °C be the final temperature.
From tables, obtain
Ca = 0.129 J/(g-°C), the specific heat of gold
Cw = 4.18 J/(g-°C), the specific heat of water
At equilibrium, heat lost by the gold - heat gained by the water.
Heat lost by the gold is
Qa = Ma*Ca*(T - Ta)
= (31.1 g)*(0.129 J/(g-°C)(*(69.3 - T °C)-
= 4.0119(69.3 - T) j
Heat gained by the water is
Qw = Mw*Cw*(T-Tw)
= (64.2 g)*(4.18 J/(g-°C))*(T - 27.8 °C)
= 268.356(T - 27.8)
Equate Qa and Qw.
268.356(T - 27.8) = 4.0119(69.3 - T)
272.3679T = 7738.32
T = 28.41 °C
Answer: 28.4 °C
Five hundred twenty million, three hundred and forty thousand. hope it helps!
Answer:
957.7mL
Explanation:
Using the formula below;
CaVa = CbVb
Where;
Ca = concentration of acid (M)
Va = volume of acid (mL)
Cb = concentration of base (M)
Vb = volume of base (mL)
According to the information provided in this question:
Ca = 0.166 M
Cb = 0.013 M
Va = 75mL
Vb = ?
Using CaVa = CbVb
0.166 × 75 = 0.013 × Vb
12.45 = 0.013Vb
Vb =12.45/0.013
Vb = 957.7mL
F = (mass)(acceleration) = ma
m = 55 kg
Vi = 20 m/s
t = 0.5 s
Vf = 0 m/s (since she was put to rest)
a=(Vf-Vi)/t
a=(0-20)/5
a = 40 m/s^2 (decelerating)
F = ma = (55 kg)(40 m/s^2)
F = 2200 N
