1. How much energy is required to be removed to lower the temperature of a bottle of drinking water (280mL) from 22°C to 4°C? The density of water is 1.0g/mL
here are the equations we can use:
(-m1)(C1)(ΔT1)=(m2)(C2)(ΔT2)
Q=mcΔT
Q=mL
There might be others I am forgetting...sorry
My teacher calls this a mixing problem because we have two substances. I just don't understand what equations to use (I am guessing this is a two or more step problem) and then where I put all the variables. Please indicate which equation you use and why you used it. Thanks!
To calculate the amount of energy required to lower the temperature of the water, we can use the specific heat capacity formula: Q = mcΔT. Here's how we can apply it to your problem step by step:
1. Identify the variables:
- Q: The amount of energy required to lower the temperature of the water in Joules (J).
- m: The mass of the water in grams (g) or kilograms (kg).
- c: The specific heat capacity of water, which is 4.18 J/g°C or 4,180 J/kg°C.
- ΔT: The change in temperature, calculated as the final temperature minus the initial temperature.
2. Convert the volume of water to mass:
- The density of water is given as 1.0 g/mL.
- The volume of water is 280 mL.
- Use the formula: mass = volume × density, which gives us mass = 280 g.
3. Calculate the change in temperature:
- The initial temperature is given as 22°C.
- The final temperature is given as 4°C.
- Calculate ΔT as 4°C - 22°C = -18°C.
4. Substitute the values into the specific heat capacity formula:
- Q = mcΔT
- Q = (mass) × (specific heat capacity) × (change in temperature)
- Q = (280 g) × (4.18 J/g°C) × (-18°C)
- Q = -201,888 J
Hence, it would require -201,888 Joules of energy to lower the temperature of the 280 mL bottle of drinking water from 22°C to 4°C. Note that the negative sign indicates that energy is being removed or released from the water as heat.
To solve this problem, we can use the equation Q = mcΔT, where Q is the amount of energy required, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
In this case, we need to calculate the energy required to lower the temperature of 280 mL (or 280 g) of water from 22°C to 4°C.
Step 1: Calculate the mass of water (m)
Since the density of water is 1.0 g/mL, the mass of 280 mL of water is 280 g.
Step 2: Calculate the change in temperature (ΔT)
The change in temperature is the final temperature minus the initial temperature. In this case, it is 4°C - 22°C = -18°C (note that we use -18°C instead of 18°C to indicate a decrease in temperature).
Step 3: Determine the specific heat capacity of water (c)
The specific heat capacity of water is approximately 4.18 J/g°C.
Step 4: Calculate the amount of energy required (Q)
Using the equation Q = mcΔT, we insert the values:
Q = (280 g)(4.18 J/g°C)(-18°C)
Q ≈ -21151.2 J
The negative sign indicates that energy is being removed from the water. Therefore, approximately 21,151.2 J of energy needs to be removed to lower the temperature of the bottle of drinking water from 22°C to 4°C.