The specific heat of granite is 0.80 kJ/kg·°C. If 2.0 MJ of heat are added to a 100−kg granite statue of James Prescott Joule that is originally at 18°C, what is the final temperature of the statue?
800 Joules / kg deg C
What is MJ ? I am familiar with mJ which is .001 Joules
Do you mean Megajoule maybe 2.0 * 10^6 Joules?
if so then
2.0 *10^6 = 800 * 100 * (T-18)
2.0*10^6 = 8 *10^4 (T-18)
25 = T-18
T = (25 + 18) deg C
q = mass x specific heat x (Tfinal-Tinitial)
q = 2E6 joules
mass = 100 kg
specific heat = 800 J/kg*C
Tfinal = ?
Tinitial = 18 C
Post your work if you need additional assistance.
Substitute and solve for Tfinal
To find the final temperature of the statue, we can use the formula:
Q = mcΔT
where:
Q = heat energy added (in joules)
m = mass of the object (in kilograms)
c = specific heat capacity (in joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's convert the given values to the appropriate units:
Specific heat of granite = 0.80 kJ/kg·°C = 0.80 × 1000 J/kg·°C = 800 J/kg·°C
Heat energy added = 2.0 MJ = 2.0 × 10^6 J
Mass of the statue = 100 kg
Initial temperature = 18°C
Now, let's rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
ΔT = (2.0 × 10^6 J) / (100 kg × 800 J/kg·°C)
Calculating:
ΔT = 25°C
Finally, we can determine the final temperature by adding the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 18°C + 25°C = 43°C
Therefore, the final temperature of the statue is 43°C.
To find the final temperature of the granite statue, we can use the formula:
Q = mcΔT
Where:
Q is the amount of heat added (in joules),
m is the mass of the substance (in kilograms),
c is the specific heat capacity of the substance (in joules per kilogram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
Given:
- Amount of heat added: 2.0 MJ = 2.0 × 10^6 J
- Mass of the statue: 100 kg
- Specific heat capacity of granite: 0.80 kJ/kg·°C = 0.80 × 10^3 J/kg·°C
- Initial temperature of the statue: 18°C
Let's substitute the given values into the formula and solve for the final temperature.
2.0 × 10^6 J = (100 kg) × (0.80 × 10^3 J/kg·°C) × ΔT
Divide both sides of the equation by (100 kg) × (0.80 × 10^3 J/kg·°C):
2.0 × 10^6 J / ((100 kg) × (0.80 × 10^3 J/kg·°C)) = ΔT
ΔT ≈ 25°C
Therefore, the final temperature of the granite statue will be 18°C + 25°C = 43°C.