· 1 J/(kg K) = 2.389x10-4 kcal/(kg oC) = 2.389x10-4 Btu/(lbm oF) · 1 kJ/(kg K) = 0.2389 kcal/(kg oC) = 0.2389 Btu/(lbm oF) · 1 Btu/(lbm oF) = 4,186.8 J/ (kg K) = 1 kcal/(kg oC) · 1 kcal/(kg oC) = 4,186.8 J/ (kg K) = 1 Btu/(lbm oF) Metal Specific Heat Capacity - cp (kJ/kg K)(kcal/kgoC) (Btu/lbmoF) Lead 0.13 0.031 0.03 · 1 J/(kg K) = 2.389x10-4 kcal/(kg oC) = 2.389x10-4 Btu/(lbm oF) · 1 kJ/(kg K) = 0.2389 kcal/(kg oC) = 0.2389 Btu/(lbm oF) · 1 Btu/(lbm oF) = 4,186.8 J/ (kg K) = 1 kcal/(kg oC) · 1 kcal/(kg oC) = 4,186.8 J/ (kg K) = 1 Btu/(lbm oF) Metal Specific Heat Capacity - cp (kJ/kg K)(kcal/kgoC) (Btu/lbmoF) Lead 0.13 0.031 0.03
The specific heat of talc is approximately 0.86 Joules per gram per degree Celsius (J/g°C).
Water has much higher specific heat than lead. All metals have fairly low specific heat values.
The specific heat capacity of tin is 0.227 J/g°C. Using the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, the heat energy required would be approximately 920.1 Joules.
This value is 198 J/kg.K at the melting point of uranium.
Water has a specific heat of 4.18J/gC, so set up the specific heat equation: C (spec. heat)=q (joules)/mass x temp. change, so: 4.18 (spec. heat of water) = q/46g(100deg), so q = 4.18(4,600) = 19,228 joules (or 19.228 kJ).
The unit for specific heat is Joules/g-Kelvin or it can be Joules/g-Celsius J= Joules g= Grams C= Celsius
The specific heat of water is 4186 joules per kilogram degree Celsius.
The specific heat capacity of lead is approximately 0.128 J/g°C. The temperature change is 85.6°C - 25.5°C = 60.1°C. Using the formula Q = mass x specific heat x temperature change, the heat required is 69g x 0.128 J/g°C x 60.1°C = 530.88 Joules.
it is in joules. 03o
it is in joules. 03o
Joule/kilogram-kelvin The SI unit is joules / kelvin. This is valid for an object of any size, but if you want the typical specific heat for a certain type of material, you have to standardize it, resulting in either joules / (kelvin x kilogram) or joules / (kelvin x mole).
To calculate the heat released, you can use the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity (0.128 J/g°C for lead), and ΔT is the temperature change. First, convert the mass to kilograms (85.0 g = 0.085 kg) and then calculate Q using the specific heat capacity of lead to get the answer in joules. Finally, convert joules to kilojoules (1 kJ = 1000 J) to find the heat released.
In SI, specific heat capacity is measured in joules per kilogram kelvin.
The unit for specific latent heat is J Kg-1(Joules per Kilogram)
The specific heat capacity of argon is 0.520 joules per gram per degree Celsius.
Specific heat of lead = 0.160 J/gC 45 pounds (454 grams/ 1 pound) = 20430 grams Use. q(Joules) = mass * specific heat * change in temp. q = (20430 grams)(0.160 J/gC)(100 C - 62 C) = 1.2 X 10^5 Joules ------------------------- or ------------------------- 1.2 X 10^5 Joules (1 calorie/4.184 Joules) = 29688 calories
The Specific Heat Capacity of water is 4,184 Joules per kg per Kelvin