If 0.045 kg of ice at 0 C is added to 0.390 kg of water at 34 C in a 0.150 kg aluminum calorimeter cup, what is the final temperature of the water?
Please help!
the sum of heats gained is zero.
heatgainedicemelting+ heatgained by water heating+ heat gained alumin heating=0
.045*Hfice+(.045)specheatwater*(Tf-0)+.390*specheatwater*(Tf-34)+.150specheatAl(Tf-34)=0
solve for Tf
is Hfice heat fushion of ice? or is latent heat the same as heat fushion?
so i did
(0.45*333.55) + (0.045*4.816)(T-0) + (0.39*4.186)(T-34) + (.15*920)(T-34) = 0
15.009 + 0.18837T + 55.51-1.63254T + 4692 - 138T = 0
4762.519 - 139.4T = 0
T = 34.16 C
But I'm getting the wrong answer. Am I using the wrong number for Al? Or Hf?
To find the final temperature of the water, we need to use the principle of conservation of energy. The heat lost by the water and ice must be equal to the heat gained by the calorimeter and any temperature change in the water.
The equation we can use to solve this problem is:
Heat gained by water and calorimeter = Heat lost by ice
The heat gained by the water and calorimeter is given by:
Q = mcΔT
Where:
Q is the heat gained
m is the mass of the water and calorimeter
c is the specific heat capacity of water
ΔT is the change in temperature
The heat lost by the ice is given by:
Q = mlf
Where:
Q is the heat lost
m is the mass of the ice
lf is the latent heat of fusion of water (amount of heat required to convert 1 kg of ice at 0°C to water at 0°C without any temperature change)
Given:
mass of ice (m1) = 0.045 kg
final temperature of water (Tf) = ? (what we need to find)
mass of water (m2) = 0.390 kg
initial temperature of water (T1) = 34°C
mass of calorimeter (m3) = 0.150 kg
specific heat capacity of water (c) = 4186 J/(kg·°C)
latent heat of fusion of water (lf) = 334000 J/kg
First, let's calculate the heat lost by the ice using:
Q = mlf
Q = (0.045 kg) × (334000 J/kg) = 15030 J
Next, let's calculate the heat gained by the water and calorimeter. We need to consider two parts: the heat required to raise the temperature of the water to the final temperature and the heat required to raise the temperature of the calorimeter.
Q1 = mcΔT
Q1 = (0.390 kg) × (4186 J/(kg·°C)) × (Tf - 34°C)
Q2 = mcΔT
Q2 = (0.150 kg) × (4186 J/(kg·°C)) × (Tf - 34°C)
Since the total heat gained must be equal to the heat lost, we can write:
Q1 + Q2 = Q
(0.390 kg) × (4186 J/(kg·°C)) × (Tf - 34°C) + (0.150 kg) × (4186 J/(kg·°C)) × (Tf - 34°C) = 15030 J
Now, we can solve this equation to find the final temperature (Tf) of the water:
(0.390 kg + 0.150 kg) × (4186 J/(kg·°C)) × (Tf - 34°C) = 15030 J
0.54 kg × (4186 J/(kg·°C)) × (Tf - 34°C) = 15030 J
2235.24 J/(kg·°C) × (Tf - 34°C) = 15030 J
Tf - 34°C = 15030 J / (2235.24 J/(kg·°C))
Tf - 34°C = 6.72 °C
Tf = 6.72 °C + 34°C
Tf ≈ 40.72 °C
Therefore, the final temperature of the water is approximately 40.72 °C.