A total of 54.0 joules of heat are absorbed as 58.3 g of lead is heated from 12.0°C to 42.0°C. From these data, what is the specific heat of lead?
To find the specific heat of lead, we can use the equation:
q = mcΔT
Where q is the heat absorbed, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature.
Given data:
q = 54.0 J
m = 58.3 g
ΔT = 42.0°C - 12.0°C = 30.0°C
Plugging these values into the equation:
54.0 J = (58.3 g)(c)(30.0°C)
We can solve for c by dividing both sides of the equation by (58.3 g)(30.0°C):
c = 54.0 J / (58.3 g)(30.0°C)
c = 54.0 J / (1749 g°C)
c ≈ 0.0309 J/g°C
Therefore, the specific heat of lead is approximately 0.0309 J/g°C.
To find the specific heat of lead, we can use the formula:
Q = m * c * ΔT
Where:
Q = Amount of heat absorbed (in joules)
m = Mass of the substance (in grams)
c = Specific heat of the substance (in J/g°C)
ΔT = Change in temperature (in °C)
In this case, we have:
Q = 54.0 J
m = 58.3 g
ΔT = 42.0°C - 12.0°C = 30.0°C
Rearranging the formula, we have:
c = Q / (m * ΔT)
Substituting the given values:
c = 54.0 J / (58.3 g * 30.0°C)
c = 54.0 J / (1749 g°C)
c ≈ 0.0309 J/g°C
Therefore, the specific heat of lead is approximately 0.0309 J/g°C.
To find the specific heat of lead, we can use the formula:
q = m * c * ΔT
Where:
q is the heat absorbed or released
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we're given that 54.0 joules of heat (q) are absorbed as 58.3 g of lead (m) is heated from 12.0°C to 42.0°C (ΔT).
So we can rearrange the formula to solve for c:
c = q / (m * ΔT)
Let's plug in the values:
c = 54.0 J / (58.3 g * (42.0°C - 12.0°C))
Now let's calculate the specific heat:
c = 54.0 J / (58.3 g * 30.0°C)
c = 54.0 J / 1749 g·°C
c ≈ 0.031 J/g·°C
Therefore, the specific heat of lead is approximately 0.031 J/g·°C.