If all of the energy from a 60-watt bulb was put into 1 gram of water at room temperature for 2 seconds, how much would the temperature of the water increase?
To calculate the increase in temperature of the water, we need to use the formula for thermal energy:
Q = mcΔT
Where:
Q is the thermal energy transferred to the water,
m is the mass of the water (1 gram),
c is the specific heat capacity of water (approximately 4.18 J/g°C), and
ΔT is the change in temperature.
We are given that the energy transferred is from a 60-watt bulb, which means it delivers 60 joules of energy per second (since 1 watt = 1 joule/second). However, we are only considering a duration of 2 seconds. Therefore, the energy transferred to the water is:
Q = (60 joules/second) * (2 seconds) = 120 joules
Now we can substitute the values into the formula to find the change in temperature (ΔT):
120 joules = (1 gram) * (4.18 J/g°C) * ΔT
Simplifying:
120 joules = 4.18 J/°C * ΔT
ΔT = (120 joules) / (4.18 J/°C)
ΔT ≈ 28.71°C
Therefore, the temperature of the water would increase approximately 28.71 degrees Celsius.