At a certain elevation, the boiling point of water is 98.5°C. How much energy is needed to heat 35.0 mL of water to the boiling point at this elevation if the water initially was at 23.4°C?
(Cp=75.38J/(mol*C), d= 1.00g/mL)
q = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Tfinal = 98.5 C
Tinitial = 23.4 C
11.0kJ
Well, heating up water sounds like a job for a hot water bottle...or a science problem! Let's dive in, shall we?
First, let's figure out how much water we have. We're given that we have 35.0 mL of water, and we know that the density of water is 1.00 g/mL. So, the mass of water is just 35.0 g. Easy peasy!
Next, we need to find the change in temperature. The water initially starts at 23.4°C and we want to heat it to the boiling point of 98.5°C. So, the change in temperature is just 98.5°C - 23.4°C = 75.1°C.
Now, we can calculate the amount of energy needed using the formula: energy = mass x specific heat capacity x change in temperature.
Since we have the mass of water and the change in temperature, we just need the specific heat capacity. We're given that the specific heat capacity of water is 75.38 J/(mol*C), but we need it in J/(g*C).
Lucky for us, we can use the molar mass of water to convert it! The molar mass of water is approximately 18 g/mol. So, we divide the specific heat capacity by the molar mass to get the specific heat capacity in J/(g*C).
Now, we're ready to crunch some numbers. Plug in the values:
energy = 35.0 g x (75.38 J/(mol*C) / 18 g/mol) x 75.1°C
Get your calculator ready to calculate that, clown bot!
To calculate the energy needed to heat 35.0 mL of water from 23.4°C to its boiling point at a certain elevation, we can follow these steps:
Step 1: Calculate the mass of water.
Given that the density (d) of water is 1.00 g/mL and the volume (V) is 35.0 mL, we can calculate the mass (m) of water using the formula:
m = d * V
m = 1.00 g/mL * 35.0 mL
m = 35.0 g
Step 2: Calculate the moles of water.
To determine the moles of water (n), we need to use the molar mass of water, which is 18.015 g/mol. The moles can be calculated using the formula:
n = m / M
n = 35.0 g / 18.015 g/mol
n ≈ 1.942 mol
Step 3: Calculate the temperature change.
The temperature change (ΔT) is the difference between the final temperature (T₂) and the initial temperature (T₁). In this case, T₁ is 23.4°C, and T₂ is the boiling point at 98.5°C. Therefore:
ΔT = T₂ - T₁
ΔT = 98.5°C - 23.4°C
ΔT = 75.1°C
Step 4: Calculate the energy using the specific heat capacity.
The specific heat capacity (Cp) of water is provided as 75.38 J/(mol°C). We can use the formula below to calculate the energy (q) needed:
q = n * Cp * ΔT
q = 1.942 mol * 75.38 J/(mol°C) * 75.1°C
q ≈ 11079.44 J
Therefore, approximately 11079.44 J of energy is needed to heat 35.0 mL of water from 23.4°C to its boiling point at this specific elevation.