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.
To convert 3 kg of ice at 0 degrees Celsius to water at the same temperature, you would need to supply 3,360,000 joules of heat (latent heat of fusion of ice is 334,000 J/kg). This energy will help melt the ice into water at 0 degrees Celsius.
The amount of energy required to raise the temperature of 1 kg of water by 1 degree Celsius is approximately 4,186 Joules. Therefore, to raise the temperature by 2 degrees Celsius, you would need about 8,372 Joules of energy.
Heat energy is typically measured in joules (J) or calories (cal). Degrees Celsius and degrees Fahrenheit are units of temperature, not energy.
To raise the temperature of a substance, you need to calculate the heat energy using the specific heat capacity of the substance. Without knowing the specific heat capacity of the substance in question, it's not possible to determine the exact amount of energy required to raise the temperature from 30 to 45 degrees Celsius.
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.
8.200 J
The change in temperature is 21 degrees Celsius. To calculate the energy required, we use the formula: Energy = mass * specific heat * change in temperature. Plugging in the values, Energy = 1.3g * 0.131 J/g°C * 21°C = 35.247 Joules. Therefore, 35.247 Joules of energy is required to heat 1.3 grams of gold from 25°C to 46°C.
The specific heat capacity of water is 4.18 J/g°C. To calculate the energy required to raise 21 kg of water by 2 degrees Celsius, use the formula: Energy = mass x specific heat capacity x temperature change. Plugging in the values, the energy required is 21,084 Joules.
Approx 4974 Joules.
The specific heat of ice at 0 degrees Celsius is approximately 2.09 Joules/gram degree Celsius. This means that it takes 2.09 Joules of energy to raise the temperature of 1 gram of ice by 1 degree Celsius.
539 calories per gram for heat of vaporization plus 1 cal/gram/degree C 100 degrees C - 80 degrees C = 20 degrees C (539 calories + 20 calories) X 50 kg X 1000 gm/kg = 27950000 cal = 27,950 kcal