7.014*10^4
one trillion pesos
The specific heat of the substance.
(110 calories) plus (heat lost from the container during the procedure)
To calculate the heat required to raise the temperature of 10 pounds of water from 50°F to a specific temperature, you can use the formula: [ Q = mc\Delta T ] where ( Q ) is the heat energy (in BTUs), ( m ) is the mass (in pounds), ( c ) is the specific heat capacity of water (approximately 1 BTU/lb°F), and ( \Delta T ) is the change in temperature (in °F). For example, if you want to raise it to 150°F, the temperature change (( \Delta T )) would be 100°F, so the heat required would be: [ Q = 10 , \text{lb} \times 1 , \text{BTU/lb°F} \times 100°F = 1000 , \text{BTUs} ] Adjust ( \Delta T ) based on your target temperature.
6 % of 1000 = 6/100 * 1000 = 0.06 * 1000 = 60
The heat required to evaporate 1 liter of water at 100 degrees Celsius is known as the latent heat of vaporization of water, which is approximately 2260 kJ/kg. Since the density of water is about 1000 kg/m³, the heat required would be around 2260 kJ.
Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal
Can you help
1,832 degrees Fahrenheit.
Assuming standard atmospheric pressure, 2260 kilojoules.
If you warm it from 35 degrees Celsius to 1000 degrees Celsius, a mas will vastly increase in volume or pressure. Without knowing how you intend to allow for that, your question is unanswerable.
To heat 1 gram of water by 1 degree Celsius, it takes 4.18 joules. So, to heat water from, for example, 20 degrees to 100 degrees, you would need to calculate the total mass of water and apply the specific heat capacity to determine the total energy required.
To calculate the energy needed to change ice at -32.9 degrees to water at 75 degrees, you need to consider the energy required for three steps: Heating ice from -32.9 degrees to 0 degrees (specific heat capacity of ice) Melting ice at 0 degrees into water at 0 degrees (latent heat of fusion of ice) Heating water from 0 degrees to 75 degrees (specific heat capacity of water) Once you have the energy needed for each step, you can add them together to find the total energy required.
The specific heat capacity of water is 4186 J/kg*C. To calculate the heat required, use the formula: heat = mass * specific heat capacity * change in temperature. Plugging in the values, the heat required to raise the temperature of 0.25 kg of water by 10 degrees Celsius is approximately 1046.5 Joules.
To calculate the heat energy required, you can use the specific heat capacity of silver, which is 0.235 J/g°C. First, convert the mass of silver to grams (1 kg = 1000 g). Then, use 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. Substituting the values, you can find the heat energy needed.
The amount of heat energy required can be calculated using the formula: Q = mcΔT. Given m = 0.362 kg, c = 390 J/kg°C for copper, and ΔT = (60.0 - 23.0) = 37.0 °C, plug these values into the formula to find the heat energy required to raise the temperature of the copper.
1000 sq ft