The specific heat capacity of lead is 0.128 J/g°C. First, calculate the change in temperature: 98.5°C - 22.0°C = 76.5°C. Next, calculate the heat energy using the formula Q = mcΔT where m = 65g, c = 0.128 J/g°C, and ΔT = 76.5°C. Substituting the values gives Q = 65g * 0.128 J/g°C * 76.5°C = 625.2 Joules.
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.
The specific heat capacity of water is 4.18 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, we can calculate that the heat required is approximately 20910 Joules or 20.91 kJ.
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.
The specific heat of water is 4.179 Joules per gram per degree Centigrade. The density of water is 1 gram per cubic centimeter, so one liter is 1000 grams. This means it takes 4179 Joules to raise one liter one degree Centigrade.
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.
Heat of vaporization of water is 2.26 x 106 joules per kg. Therefore 1 gram of water will need 2.26 x 103 joules.
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degrees and centigrade
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.
334.8 Joules
The specific heat capacity of water is 4.18 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, we can calculate that the heat required is approximately 20910 Joules or 20.91 kJ.
Heat energy is typically measured in joules (J) or calories (cal). Degrees Celsius and degrees Fahrenheit are units of temperature, not energy.
q = mass * specific heat * change in temp. ( Joules, so conversion at end ) q = (93.6 g)(2.51 J/gC)[55o C - ( - 35o C)] = 21144.24 Joules ===============================
70 calories per gram. (The specific heat capacity of water is 1 calorie per gram per degree C.) This could be converted into Joules if necessary using the conversion factor of 1 calorie = 4.18400 Joules.
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.
The specific heat capacity of water is 4.184 J/g°C. The change in temperature is 35.5°C - 21°C = 14.5°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, you can calculate that the heat required is approximately 1554.7 joules.
1.25