A 200W lamp is totally immersed in 1500gm of water. How much will the temperature of the wtaer rise in 3 mins? Neglect heat losses from the water.
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Energy= mass*specificheatwater*deltaTEmp
where energy equal watts*timeinSeconds
Ah, no need to get all Watts and specific heats with me. Let's keep it light and simple!
So, a 200W lamp completely immersed in 1500gm of water for 3 minutes... I'm afraid it won't be enough to turn the water into a hot tub. Don't expect a steamy situation!
But if we really want to crunch some numbers, the change in temperature (delta T) can be calculated using the equation:
Change in temperature = (Energy / (mass * specific heat))
Now, considering the energy supplied by the lamp, which is 200W for 3 minutes (180 seconds), and assuming the specific heat of water is 4.186 J/g°C, we can plug in the values and see what we get:
Change in temperature = (200W * 180s) / (1500g * 4.186 J/g°C)
Hold on to your swim trunks, the result is... well, nearly negligible. But hey, it's still something!
So, the temperature of the water will rise by approximately... drumroll, please... 0.08°C! Yes, you've got it, a whopping 0.08 degrees! That's about as exciting as watching water evaporate. Don't get too hot and bothered now! Enjoy your lukewarm water experience.
To find out how much the temperature of the water will rise, we need to use the formula:
Energy = mass * specific heat of water * change in temperature
In this case, the energy is given by the power of the lamp, which is 200 watts, multiplied by the time in seconds, which is 3 minutes or 180 seconds.
So, the energy is calculated as:
Energy = 200 watts * 180 seconds
Next, we need to calculate the mass of the water. The given mass is 1500 grams, but we need to convert it to kilograms since the specific heat of water is typically given in joules per kilogram per degree Celsius.
So, the mass of the water is:
Mass = 1500 grams / 1000 grams per kilogram
Now, we can calculate the change in temperature using the rearranged formula:
Change in temperature = Energy / (mass * specific heat of water)
The specific heat of water is approximately 4.186 joules per gram per degree Celsius.
Plugging in the values, the calculation becomes:
Change in temperature = (200 watts * 180 seconds) / (1500 grams / 1000) * 4.186 joules per gram per degree Celsius
Simplifying further:
Change in temperature = (200 * 180) / (1.5) * 4.186 joules per degree Celsius
Change in temperature = 40,048 / 1.5 * 4.186 joules per degree Celsius
Change in temperature ≈ 44,507 joules per degree Celsius
Since heat losses from the water are neglected, all the energy will be transferred to the water, resulting in a rise in temperature of approximately 44,507 degrees Celsius.
To find the change in temperature of the water, we need to calculate the energy transferred from the lamp to the water.
We know that power is given as 200W, which means it is transferring 200 joules of energy per second (since 1 watt is equal to 1 joule per second).
Since we are given the time in minutes, we need to convert it to seconds. There are 60 seconds in a minute, so 3 minutes is equal to 3 * 60 = 180 seconds.
Now we can calculate the energy transferred to the water:
Energy = Power * Time
= 200W * 180s
= 36000 joules
Now, we need to calculate the change in temperature of the water. We can use the specific heat capacity of water, which is approximately 4.18 joules per gram per degree Celsius (J/g°C).
Given that the mass of water is 1500 grams, we can calculate the change in temperature using the formula:
Change in Temperature = Energy / (mass * specific heat capacity)
Change in Temperature = 36000 joules / (1500 g * 4.18 J/g°C)
≈ 5.42 °C
Therefore, the temperature of the water will rise by approximately 5.42 degrees Celsius in 3 minutes. Remember that this calculation neglects any heat losses from the water.