How much heat is required to raise by 10.30°C the temperature of 1.2 kg of water?
How much heat is required to raise by 10.30°C the temperature of 1.2 kg of iron?
Q1 = c1•m1 •Δt1 =4180•1.2•10.3 = 51665 J
Q2 = c2•m2• Δt2 =444•1.2•10.3 = 5488 J
Calculate the heat required to raise the temperate of 10kg of copper perform from 20c to 30c specific capacity of copper= 400j/kgk
To find the amount of heat required to raise the temperature of a substance, you can use the specific heat capacity formula:
Q = mcΔT
Q: Heat energy (in Joules)
m: Mass of the substance (in kilograms)
c: Specific heat capacity of the substance (in Joules per kilogram per degree Celsius)
ΔT: Change in temperature (in degrees Celsius or Kelvin)
For water:
m = 1.2 kg
c = 4186 J/kg°C (specific heat capacity of water)
ΔT = 10.30°C
Using the formula, we can calculate the heat energy required:
Q = (1.2 kg) * (4186 J/kg°C) * (10.30°C)
= 51,066.24 J
Therefore, 51,066.24 Joules of heat energy is required to raise the temperature of 1.2 kg of water by 10.30°C.
For iron:
m = 1.2 kg
c = 450 J/kg°C (specific heat capacity of iron)
ΔT = 10.30°C
Using the formula, we can calculate the heat energy required:
Q = (1.2 kg) * (450 J/kg°C) * (10.30°C)
= 5,265 J
Therefore, 5,265 Joules of heat energy is required to raise the temperature of 1.2 kg of iron by 10.30°C.