If the specific heat of water is 4.180 J/ g °C, what mass of water can be heated from 5.0°C to 80.0°C by 27000 J of heat?
The formula for calculating the amount of heat transferred to a substance is:
Q = m * c * ΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature of the substance. In this problem, we are given the specific heat of water and the amount of heat transferred, and we need to find the mass of water.
First, we can rearrange the formula to solve for the mass:
m = Q / (c * ΔT)
Substituting the given values, we get:
m = 27000 J / (4.180 J/g°C * (80°C - 5°C))
m = 27000 J / (4.180 J/g°C * 75°C)
m = 90.6 g
Therefore, a mass of 90.6 grams of water can be heated from 5.0°C to 80.0°C by 27000 J of heat.
6.00 g of gold was heated from 20.0oC to 22.0oC. How much heat was applied to the gold? Specific heat of gold is 0.129 J/g°C.
The formula for calculating the amount of heat transferred is:
Q = m * c * ΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat of the substance, and ΔT is the change in temperature of the substance.
Substituting the given values, we get:
Q = 6.00 g * 0.129 J/g°C * (22.0°C - 20.0°C)
Q = 6.00 g * 0.129 J/g°C * 2.0°C
Q = 1.55 J
Therefore, 1.55 J of heat was applied to the gold.
A 2.80 g piece of metal absorbs 14.0 J of energy as it is heated from 22.0mc006-2.jpgC to 35.0mc006-3.jpgC. Find the specific heat of this metal and identify the metal.
The formula to calculate the specific heat of a substance is:
c = Q / (m * ΔT)
where c is the specific heat, Q is the amount of heat absorbed or released, m is the mass of the substance, and ΔT is the change in temperature.
Substituting the given values, we get:
c = 14.0 J / (2.80 g * (35.0°C - 22.0°C))
c = 14.0 J / (2.80 g * 13.0°C)
c = 0.359 J/g°C
Therefore, the specific heat of the metal is 0.359 J/g°C.
To identify the metal, we can compare this specific heat to known values for specific heat. A few common metals and their specific heats are:
- Copper: 0.385 J/g°C
- Aluminum: 0.900 J/g°C
- Lead: 0.128 J/g°C
- Iron: 0.449 J/g°C
Since the specific heat of the metal we are trying to identify is less than copper, iron, and aluminum, but greater than lead, it is likely that the metal is lead.
To find the mass of water that can be heated from 5.0°C to 80.0°C by 27000 J of heat, we can use the formula:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of water
c is the specific heat of water
ΔT is the change in temperature
In this case, we know:
Q = 27000 J
c = 4.180 J/g °C
ΔT = (80.0°C - 5.0°C) = 75.0°C
Let's rearrange the formula to solve for the mass of water (m):
m = Q / (c * ΔT)
Now we can simply substitute the known values:
m = 27000 J / (4.180 J/g °C * 75.0°C)
m = 27000 J / 313.5 J/g
m ≈ 86.18 g
Therefore, the mass of water that can be heated from 5.0°C to 80.0°C by 27000 J of heat is approximately 86.18 grams.