: ΔQ = mcΔT : ΔQ = (60g) (0.385 J g−1 K−1) (60 °C same as 60 Kelvin) : :: = 1386 Joules
q = (250 g)(0.46 J/gC)(300 C - 27 C) = 3.1 X 104 Joules -------------------------
To calculate the heat required to melt 3 x 10² kg of copper, we use the formula ( Q = m \cdot L_f ), where ( Q ) is the heat, ( m ) is the mass, and ( L_f ) is the latent heat of fusion for copper, which is approximately 200 kJ/kg. For 3 x 10² kg of copper, the heat required would be ( Q = 3 \times 10² , \text{kg} \times 200 , \text{kJ/kg} = 60,000 , \text{kJ} ). Thus, it takes about 60,000 kJ of heat to melt that amount of copper.
The nagle of light determines the area over which the energy of the light is spread out and that will affect how much it is heated.
None. They are 75% copper and 25% copper.
1963 COPPER penny is worth half a billion dollars.
The specific heat capacity of copper is 0.385 J/g°C. You can use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, you can calculate the heat absorbed by the copper.
The specific heat capacity of iron is 0.449 J/g°C. To calculate the heat absorbed, use the formula Q = mcΔT, where Q is the heat absorbed, m is the mass in grams, c is the specific heat capacity, and ΔT is the temperature change. Plugging in the values, the heat absorbed by 1.0 g of iron heated to 15°C is 6.735 J.
51%
Joule (J) is a unit of energy.Gram (g) is a unit for mass.
A copper wire is a much better conductor of heat than air is.
To find the heat energy absorbed by the lead block, you need to use the specific heat capacity of lead, which is 128 J/kg°C. The formula to calculate heat energy is: Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, you can find the heat energy absorbed by the lead block.
Solid copper appears to gain mass when heated in air, because the copper reacts with oxygen in the air to form copper oxide. However, the actual mass of the copper does not increase; the mass of the solid increases by a value equal to the mass of oxygen removed from the air.
1,386J the formula needed in q=mC Delta T Delta is usually written as a triangle, but that not available here. q is heat m is mass C is specific heat (of copper in this case) every substance has a constant delta T means change of temp, in degrees centigrade to match units of "C" so, calculate delta T, convert to centigrade first then plug in numbers, multiply and go. 353 K - 293K = 80 Centigrade - 20 Centigrade = 60 Centigrade see... q = (60.00 G)(0.39J/gxdegrees centigrade)(60 C) all units cancel out except joules, the unit of heat, 1404 Joules 1386!!
It depends how much heat is applied and the mass of the gold.
all of them, just depends on how much heat you use
It depend upon the temperature how much you heat it
q = mHvq = heatm = mass (30g)Hv = heat of vaporization (2,260J/g)q = (30g)(2,260J/g)q = 67,800JWhen 30 grams of water is converted into steam, how much heat is absorbed?67,800J of heat, also represented as 67.8kJ of heat is absorbed.