q( in Joules ) = mass * specific heat * change in temperature [ convert temps--Tf = Tc(1.80) + 32 ]
q = (40 g)(0.90 J/gC)(61.1o C - 22.8o C)
= 1.4 X 103 Joules
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Heat energy Q = mass x specific heat capacity x temperature change. Q = m*c*delta T Q = Joules m = kg c (aluminum) = 895.8 J/kg delta T = degr.C temp. change Answer: Q = (20/1000) x 895.8 x 5 = 89.58 Joules (Specific heat capacity of aluminum is obtained by multiplying its specific heat of 0.214 with c of water which is 4186 J/kg = 0.214 x 4186 = 895.8 J/kg).
The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.
If you want to be pedantic, scientists measure temperature in kelvins, not degrees. Heat is energy and is measured in energy units, like joules.
q(joules) = mass * specific heat * change in temperature ( 8 kg = 8000 grams ) q = (8000 grams H2O)(4.180 J/gC)(70o C - 20o C) = 1.7 X 106 joules ============
Specific heat for aluminium = 0.214 Heat required = 38.2 x 0.214 x (275 - 102) = 1414.24 calories
Approx 4974 Joules.
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.
Heat energy is typically measured in joules (J) or calories (cal). Degrees Celsius and degrees Fahrenheit are units of temperature, not energy.
The change in temperature is 25 degrees Celsius, meaning it takes 22.48 joules per degree of change. The specific heat of iron is 0.449 J/g degree Celsius. This means that the mass of iron must be 50.07 grams
Heat energy Q = mass x specific heat capacity x temperature change. Q = m*c*delta T Q = Joules m = kg c (aluminum) = 895.8 J/kg delta T = degr.C temp. change Answer: Q = (20/1000) x 895.8 x 5 = 89.58 Joules (Specific heat capacity of aluminum is obtained by multiplying its specific heat of 0.214 with c of water which is 4186 J/kg = 0.214 x 4186 = 895.8 J/kg).
How much heat (in calories) is required to heat a 43 g sample of aluminum from 72 F to 145F
The specific heat capacity of aluminum is 0.897 J/g°C. To calculate the energy required to heat 0.5kg of aluminum by a certain temperature change, you would use the formula: Energy = mass x specific heat capacity x temperature change If you have the temperature change, you can plug the values into the formula to find the total energy in joules.
46 calories (or 192, 464 joules) for each Celsius degree.
The latent heat of condensation of steam is 2260 Joules per gram (539.3 cals/g). So the amount of heat released by 12.4 g = 12.4*2260 Joules = 28,024 Joules or 6687 cals.
If you want to be pedantic, scientists measure temperature in kelvins, not degrees. Heat is energy and is measured in energy units, like joules.
The heat resistance of aluminum foil tape is typically around 600 degrees Fahrenheit.
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.