A COPPER BLOCK ,MASS 500g AND AT 200 DEGREES CELCIUS GIVES OUT ENERGY AT AN AVERAGE RATE OF 50W.
(i)what is temperature after 5 min ?
(ii)how long will it take before its temperature drops to 50 degrees celcius
POwer= heat/time= mass*specificheat*(tf-ti)/time where time is in seconds.
i) change time to seconds, solve for tf.
ii) solve for time.
To solve this problem, we can use the equation for heat transfer:
Q = mc∆T
Where:
Q = Heat transferred (in Joules)
m = Mass of the copper block (in kg)
c = Specific heat capacity of copper (in J/kg°C)
∆T = Change in temperature (in °C)
First, let's find the heat transferred in 5 minutes (convert to seconds, since power is given in Watts):
T = 5 minutes = 5 * 60 = 300 seconds
Power = 50 Watts
Heat transferred = Power * Time
Q = Pt = 50 * 300 = 15000 J
Now, we can rearrange the equation to find the change in temperature (∆T):
∆T = Q / (mc)
Using the specific heat capacity of copper (c = 387 J/kg°C) and the mass of the block (m = 500 g = 0.5 kg), we can substitute these values into the equation:
∆T = 15000 / (0.5 * 387) = 77.32 °C
(i) The temperature after 5 minutes will be the initial temperature of 200 °C minus the change in temperature:
Temperature after 5 minutes = 200 - 77.32 = 122.68 °C
(ii) To find the time it takes before the temperature drops to 50 °C, we need to rearrange the equation and solve for time:
∆T = Q / (mc)
∆T = (final temperature - initial temperature)
500 = 50 * t / (0.5 * 387)
250 / (0.5 * 387) = t
t ≈ 1.29 seconds
Therefore, it will take approximately 1.29 seconds for the temperature to drop from 200 °C to 50 °C.