how much heat must be added to 300g of water to raise its temperature from 20 to 80 degrees celcius?
q = mass x specific heat x delta T.
Note the correct spelling of celsius.
A calorimeter of heat capacity 48 JK-1 containing water at 20oC. A 200W electric heater is used to raise the water to boiling point of 100oC in 4 minutes; it is found that after a further 6 minutes, 0.03 kg of water has been evaporated. Calculate
(i) The amount of water in the calorimeter before boiling (neglecting any loss due to evaporation)
(ii) The specific latent heat of steam. (Specific heat capacity of copper = 400 J kg-1K-1)
What mass
To find out how much heat must be added to raise the temperature of water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the heat energy (in Joules),
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (in °C).
In this case, the mass of water (m) is given as 300 grams, the specific heat capacity of water (c) is 4.18 J/g°C, and the change in temperature (ΔT) is from 20 to 80 degrees Celsius.
So, plugging in the values:
Q = 300g * 4.18 J/g°C * (80°C - 20°C)
Q = 300g * 4.18 J/g°C * 60°C
Q = 75240 J
Therefore, to raise the temperature of 300g of water from 20 to 80 degrees Celsius, approximately 75240 Joules of heat must be added.