How many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C?
so its setup like
(20*1.00)*(90-50)+(20*1.00)*(90-50) = 1600g?
How many grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C?
To solve this problem, we can use the formula:
Q = mcΔT
where Q is the amount of heat transferred, m is the mass of the water, c is the specific heat capacity of water (1.00 cal/g°C), and ΔT is the change in temperature.
First, let's calculate the amount of heat lost by the 800g of water at 90°C as it cools to 50°C:
Q1 = (800g)(1.00 cal/g°C)(90°C - 50°C)
Q1 = 32000 cal
Next, let's calculate the amount of heat gained by the unknown mass of water at 20°C as it heats up to 50°C:
Q2 = (m)(1.00 cal/g°C)(50°C - 20°C)
Q2 = 30m cal
Since the heat lost by the hot water equals the heat gained by the cold water (assuming no heat loss to the surroundings), we can set Q1 equal to Q2 and solve for m:
32000 cal = 30m cal
m = 1066.67 g
Therefore, we need 1066.67 grams of water at 20°C to change 800g of water at 90°C to 50°C.
Does the distance between minima increase or decrease when the slit width is increased?
The distance between the minima in a double-slit interference pattern (also known as the interference fringe spacing) decreases as the slit width is increased. This is because a wider slit allows more light to pass through and results in a broader diffraction pattern, leading to a smaller fringe spacing. This phenomenon can be explained by the equation for the fringe spacing:
d = λD/d
where d is the slit width, λ is the wavelength of light, D is the distance between the slit and the screen, and d is the distance between adjacent fringes (minima or maxima). As the slit width increases, d decreases, resulting in a smaller distance between the fringes.
To solve this problem, we need to use the principle of conservation of energy. The formula for calculating the heat gained or lost by a substance is given by:
Q = m * c * ΔT
Where:
Q is the heat gained or lost by the substance.
m is the mass of the substance.
c is the specific heat capacity of the substance.
ΔT is the change in temperature of the substance.
In this case, we want to find the mass of water needed to change the temperature of 800g of water from 90°C to 50°C. We can set up the equation as follows:
Q1 + Q2 = 0
The initial temperature of the water is 90°C, so the heat lost by the water is:
Q1 = m1 * c * ΔT1
Where:
m1 is the mass of the water at 90°C.
ΔT1 is the change in temperature from 90°C to the final temperature.
Similarly, the heat gained by the water at 20°C is:
Q2 = m2 * c * ΔT2
Where:
m2 is the mass of the water at 20°C.
ΔT2 is the change in temperature from 20°C to the final temperature.
Since the heat lost by the water is equal to the heat gained by the water, we can rewrite the equation as:
m1 * c * ΔT1 + m2 * c * ΔT2 = 0
Substituting the given values:
800g * 1.00 cal/g°C * (90°C - 50°C) + m2 * 1.00 cal/g°C * (20°C - 50°C) = 0
Simplifying:
800 * 40 + m2 * (-30) = 0
32000 - 30m2 = 0
30m2 = 32000
m2 = 32000 / 30
m2 ≈ 1066.67g
So, approximately 1066.67 grams of water at 20°C are necessary to change 800g of water at 90°C to 50°C.