200 kg of steel. Steel is denser than aluminum (so, more stuff to heat in 200 kg of steel than in 200 kg of aluminum), and 200 kg is twice as much as 100 kg (so, more stuff to heat in 200 kg of steel than in 100 kg.)
90,000
The specific heat capacity of aluminum is 0.9 J/g°C. To calculate the energy required to raise the temperature of 0.2kg of aluminum by 3 degrees Celsius, you would use the formula: Energy = mass x specific heat capacity x temperature change. Substituting the values into the formula, Energy = 0.2kg x 0.9 J/g°C x 3°C = 0.54 Joules.
To bring the ice block to 0 degrees Celsius, you would need 150,000 Joules (Q = mcΔT). To melt the ice at 0 degrees Celsius, you would need 3,375,000 Joules (Q = mLf). Heating the water from 0 to 100 degrees Celsius would require 1,500,000 Joules (Q = mcΔT). Turning the water to steam at 100 degrees Celsius would need 10,500,000 Joules (Q = mLv). Finally, heating the steam to 120 degrees Celsius would require 600,000 Joules (Q = mcΔT). In total, you would need 15,125,000 Joules of heat energy.
The specific heat capacity of aluminum is 900 J/kg°C. The change in temperature is 3°C. Using the formula Q = mcΔT, the energy required would be 0.2 kg * 900 J/kg°C * 3°C = 540 J. So, the energy required to raise the temperature is 540 Joules.
Heat energy is typically measured in joules (J) or calories (cal). Degrees Celsius and degrees Fahrenheit are units of temperature, not energy.
25degres celsius has more thermal energy
False. Heat is measured in units of energy such as calories or joules, not in degrees Celsius. Temperature, on the other hand, is measured in degrees Celsius.
80cal/g
Thermal energy is a measure of the internal energy of a system and is not dependent on the unit of temperature used. Therefore, the thermal energy at 0 degrees Celsius would be the same as at 48 degrees Fahrenheit.
The heat energy required can be calculated using the formula: Q = mcΔT, where Q is the heat energy, m is the mass of the aluminum (0.055 kg), c is the specific heat capacity of aluminum (900 J/kg°C), and ΔT is the change in temperature (94.6°C - 22.4°C = 72.2°C). Plugging in the values, we get Q = 0.055 kg * 900 J/kg°C * 72.2°C = 3582.7 J. Hence, 3582.7 Joules of heat energy is needed to raise the temperature of the aluminum sample.
Absolute zero is -273.15 degrees Celsius. At this temperature, particles have minimum movement and energy.
The specific heat capacity of aluminum is 0.902 J/g°C. First, convert 3kg to grams (3000g). Then, calculate the change in temperature (23°C - 18°C = 5°C). Finally, use the formula Q = mcΔT to find the energy required: Q = 3000g * 0.902 J/g°C * 5°C.