An unknown mass of each of the following substances, initially at 23.0C, absorbs 1910 J of heat. The final temperature is recorded as indicated. Find the mass of each substance.
sand (Tf=62.0C)
ethanol (Tf=44.1C)
water (Tf=32.1C)
q = mass substance x specific heat x (Tfinal-Tinitial)
To find the mass of each substance, we'll use the equation:
q = mcΔT
where:
q = heat absorbed/released (in joules)
m = mass
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
Given information:
Initial temperature (Ti) = 23.0°C
Final temperature (Tf) = given for each substance
Heat absorbed (q) = 1910 J
Let's find the mass of each substance one by one.
1. Sand with Tf = 62.0°C:
Using the equation q = mcΔT, we have:
1910 = m × c × (Tf - Ti)
1910 = m × c × (62.0 - 23.0)
1910 = m × c × 39.0
The specific heat capacity of sand is approximately 0.20 J/g°C. Substituting this value, we get:
1910 = m × 0.20 × 39.0
Now, solving for m:
m = 1910 / (0.20 × 39.0)
m ≈ 24.6 grams
Therefore, the mass of sand is approximately 24.6 grams.
2. Ethanol with Tf = 44.1°C:
Using the same formula, q = mcΔT, we have:
1910 = m × c × (Tf - Ti)
1910 = m × c × (44.1 - 23.0)
1910 = m × c × 21.1
The specific heat capacity of ethanol is approximately 2.44 J/g°C. Substituting this value, we get:
1910 = m × 2.44 × 21.1
Now, solving for m:
m = 1910 / (2.44 × 21.1)
m ≈ 36.8 grams
Therefore, the mass of ethanol is approximately 36.8 grams.
3. Water with Tf = 32.1°C:
Using the same formula, q = mcΔT, we have:
1910 = m × c × (Tf - Ti)
1910 = m × c × (32.1 - 23.0)
1910 = m × c × 9.1
The specific heat capacity of water is approximately 4.18 J/g°C. Substituting this value, we get:
1910 = m × 4.18 × 9.1
Now, solving for m:
m = 1910 / (4.18 × 9.1)
m ≈ 51.6 grams
Therefore, the mass of water is approximately 51.6 grams.
To summarize:
- Mass of sand ≈ 24.6 grams
- Mass of ethanol ≈ 36.8 grams
- Mass of water ≈ 51.6 grams
To find the mass of each substance, we can use the equation for heat transfer: q = mcΔT
Where:
q = amount of heat transferred
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, we need to rearrange the equation to solve for mass:
m = q / (c * ΔT)
To use this equation, we need to know the specific heat capacity for each substance.
For sand, the specific heat capacity is typically around 0.19 J/g°C.
For ethanol, the specific heat capacity is typically around 2.44 J/g°C.
For water, the specific heat capacity is around 4.18 J/g°C.
Let's calculate the mass for each substance.
1. Sand (Tf = 62.0°C):
Given:
Initial temperature (Ti) = 23.0°C
Final temperature (Tf) = 62.0°C
Heat transferred (q) = 1910 J
Specific heat capacity (c) = 0.19 J/g°C
First, find the change in temperature:
ΔT = Tf - Ti = 62.0°C - 23.0°C = 39.0°C
Now, we can calculate the mass:
m = q / (c * ΔT) = 1910 J / (0.19 J/g°C * 39.0°C)
2. Ethanol (Tf = 44.1°C):
Given:
Initial temperature (Ti) = 23.0°C
Final temperature (Tf) = 44.1°C
Heat transferred (q) = 1910 J
Specific heat capacity (c) = 2.44 J/g°C
Again, find the change in temperature:
ΔT = Tf - Ti = 44.1°C - 23.0°C = 21.1°C
Calculate the mass:
m = q / (c * ΔT) = 1910 J / (2.44 J/g°C * 21.1°C)
3. Water (Tf = 32.1°C):
Given:
Initial temperature (Ti) = 23.0°C
Final temperature (Tf) = 32.1°C
Heat transferred (q) = 1910 J
Specific heat capacity (c) = 4.18 J/g°C
Again, find the change in temperature:
ΔT = Tf - Ti = 32.1°C - 23.0°C = 9.1°C
Calculate the mass:
m = q / (c * ΔT) = 1910 J / (4.18 J/g°C * 9.1°C)
Now, plug in the values and calculate the mass for each substance.