you chop 50 g of ice at 0 degrees celsius into 200g of water at 40 degress celcius. What is the final temperature?
The ice acquires 50 g * 80 Cal/g = 4000 Cal from the water while melting, but remining 0 C. In losing that amount of heat, the original liquid water cools by 4000 Cal/(200g*1 Cal/deg C) = 20 C, leviong it at 20 C.
Now you mix 200 g of water at 20 C with 50 g at 0 C.
At equilibrium T,
50*T = 200*(20 - T)
250 T = 4000
T = 16 C
To find the final temperature after mixing the ice and water, we can use the principle of conservation of energy.
First, let's calculate the heat gained by the ice (Q1) and the heat gained by the water (Q2):
Q1 = mass × specific heat capacity × change in temperature
Q1 = 50 g × 2.09 J/g°C × (0°C - final temperature)
Q2 = mass × specific heat capacity × change in temperature
Q2 = 200 g × 4.18 J/g°C × (final temperature - 40°C)
Since there is no heat lost or gained during the process (assuming an isolated system), the heat lost by the ice is equal to the heat gained by the water:
Q1 = Q2
Substituting the equations, we get:
50 g × 2.09 J/g°C × (0°C - final temperature) = 200 g × 4.18 J/g°C × (final temperature - 40°C)
Now we can solve this equation to find the final temperature:
50 × 2.09 × (0 - final temperature) = 200 × 4.18 × (final temperature - 40)
104.5 × final temperature = 836 × (final temperature - 40)
104.5 × final temperature = 836 × final temperature - 33440
836 × final temperature - 104.5 × final temperature = 33440
731.5 × final temperature = 33440
final temperature = 33440 / 731.5
final temperature ≈ 45.69°C
Therefore, the final temperature after mixing the ice and water would be approximately 45.69 degrees Celsius.