My Chem professor is out for the weekend and I'm confused about what the question is asking or the equations needed...
A 10.0-g block of pure nickel
(Cp = 0.445 J/g-°C) is brought to each of the temperatures below. A sealed beaker containing 15.0 g water at the temperature listed is placed on top of the nickel block and the system is allowed to equilibrate. Calculate the final temperature of the water and the state(s) at which it exists for each scenario. You may assume no heat is lost to the beaker or the surroundings. (Also assume that the average heat capacity for liquid water is 4.180 J/g-°C).
(Four parts, a-d).
(a) initial temperature of nickel = 82.1°C; initial water sample = ice at
−5.52°C
The final temperature of the water is ____ degrees Celsius.
Thanks for your help!!
Why did the nickel block and the water go on a date?
Because they wanted to heat things up!
To solve this problem, we can use the principle of energy conservation, which states that the heat gained by the water is equal to the heat lost by the nickel block.
First, let's calculate the amount of heat lost by the nickel block:
Heat lost by nickel (q1) = mass of nickel (m1) * specific heat capacity of nickel (Cp1) * change in temperature of nickel (ΔT1)
q1 = 10.0 g * 0.445 J/g-°C * (T1 - 82.1°C)
Next, let's calculate the amount of heat gained by the water:
Heat gained by water (q2) = mass of water (m2) * specific heat capacity of water (Cp2) * change in temperature of water (ΔT2)
q2 = 15.0 g * 4.180 J/g-°C * (T2 - (-5.52°C))
Since no heat is lost to the surroundings, q1 must be equal to q2:
10.0 * 0.445 * (T1 - 82.1) = 15.0 * 4.180 * (T2 - (-5.52))
Simplifying this equation, we can solve for T2, which represents the final temperature of the water:
10.0 * 0.445 * T1 - 10.0 * 0.445 * 82.1 = 15.0 * 4.180 * T2 - 15.0 * 4.180 * (-5.52)
Now, plug in the given values for T1 and solve for T2:
10.0 * 0.445 * 82.1 = 15.0 * 4.180 * T2 - 15.0 * 4.180 * (-5.52)
After solving this equation, you should find that the final temperature of the water (T2) is a positive value. This represents the final temperature of the water in degrees Celsius.
To calculate the final temperature of the water in this scenario, you need to use the principle of energy conservation and apply the formula for heat transfer.
The heat lost by the nickel block is equal to the heat gained by the water. The equation for heat transfer is:
q = m × Cp × ΔT
Where:
q = heat gained or lost (in this case, it is the heat gained by water)
m = mass of the substance (water in this case)
Cp = specific heat capacity of the substance (water in this case)
ΔT = change in temperature
Let's break down the process step by step:
Step 1: Calculate the heat lost by the nickel block.
q_nickel = m_nickel × Cp_nickel × ΔT_nickel
Given:
m_nickel = 10.0 g
Cp_nickel = 0.445 J/g-°C
ΔT_nickel = final temperature of nickel - initial temperature of nickel
Step 2: Calculate the heat gained by the water.
q_water = m_water × Cp_water × ΔT_water
Given:
m_water = 15.0 g
Cp_water = 4.180 J/g-°C
ΔT_water = final temperature of water - initial temperature of water
Step 3: Set up the equation using the principle of energy conservation:
q_nickel = q_water
Step 4: Calculate the final temperature of the water:
q_nickel = q_water
m_nickel × Cp_nickel × ΔT_nickel = m_water × Cp_water × ΔT_water
Now, you need to solve for the final temperature of the water, which is ΔT_water. Rearrange the equation to isolate ΔT_water:
ΔT_water = (m_nickel × Cp_nickel × ΔT_nickel) / (m_water × Cp_water)
Finally, calculate the final temperature of the water by adding the initial temperature of the water to ΔT_water:
Final temperature of water = initial temperature of water + ΔT_water
Plugging in the given values:
Initial temperature of water = -5.52 °C
Initial temperature of nickel = 82.1 °C
Substitute these values into the equations to get the final temperature of the water. Remember to convert all temperatures to the same unit (in this case, °C).
I hope this explanation helps you understand how to approach the problem and find the final temperature of the water.