Monique has 250 grams of water in a blue insulated cup at 15 °C. In a red insulated cup, she has some water that is at 75 °C. When she mixes the cold water into the warm water, the temperature of all the water reaches 35 °C. (The specific heat of water is 4.84 J g^−1 °C^−1.)
a) Write an algebraic equation that represents the heat gained by the cold water. Write the number equation (with units!) for the heat gained by the cold water. Calculate the heat.
b) Write an algebraic equation that represents the heat lost by the cold water. Write the number equation (with units!) for the heat lost by the cold water. Calculate the heat.
c) How are the heat gained by the cold water and the heat lost by the warm water related? Answer with words and a math equation.
d) Determine the mass of the warm water initially in the red cup.
Gained by cold = (4.84 J g^−1 °C^−1)( 250 g)(35 °C - 15 °C)
= 24,200 Joules
Lost by hot (you mean) = (4.84 J g^−1 °C^−1) (m grams) (75 °C - 35 °C)
Lost by hot = gained by cold if insulation is perfect
24,200 Joules = (4.84 J g^−1 °C^−1) (m grams) (75 °C - 35 °C)
m = [24,200 /( 4.84 * 40)] grams
= 125 grams which you could have said at the start because its temperature change is twice as much as the cold so it needs half the mass.
thanks so much Damon :D
a) To find the heat gained by the cold water, we can use the equation:
Heat gained = mass * specific heat * temperature change
Let's assign variables for the unknowns:
mass of cold water = m1 (given as 250 grams)
specific heat of water = c (given as 4.84 J g^−1 °C^−1)
initial temperature of cold water = T1 (given as 15 °C)
final temperature of mixed water = T2 (given as 35 °C)
The algebraic equation representing the heat gained by the cold water is:
Heat gained = m1 * c * (T2 - T1)
Substituting the given values, we have:
Heat gained = 250 * 4.84 * (35 - 15)
Now, we can calculate the heat gained by the cold water.
b) To find the heat lost by the cold water, we'll use the same equation:
Heat lost = mass * specific heat * temperature change
Let's assign variables for the unknowns:
mass of warm water = m2 (unknown)
specific heat of water = c (given as 4.84 J g^−1 °C^−1)
initial temperature of warm water = T2 (given as 75 °C)
final temperature of mixed water = T2 (given as 35 °C)
The algebraic equation representing the heat lost by the cold water is:
Heat lost = m2 * c * (T2 - T2)
Simplifying the equation:
Heat lost = 0 (since the change in temperature is 0 for the warm water)
We can safely state that the heat lost by the cold water is zero.
c) The heat gained by the cold water and the heat lost by the warm water are related because they are the same value. According to the Law of Conservation of Energy, energy cannot be created or destroyed, only transfer from one form to another. Therefore, the heat gained by the cold water is equal to the heat lost by the warm water.
Mathematically, it can be represented as:
Heat gained by cold water = Heat lost by warm water
Substituting the equations:
m1 * c * (T2 - T1) = m2 * c * (T2 - T2)
m1 * c * (T2 - T1) = m2 * c * 0
m1 * c * (T2 - T1) = 0
Since the heat lost by the warm water is zero, the heat gained by the cold water is also zero.
d) To determine the mass of the warm water initially in the red cup, we can simply use the equation from part b:
Heat lost = mass * specific heat * temperature change
Since the heat lost is zero, we have:
0 = mass * c * (T2 - T2)
Simplifying the equation:
0 = mass * c * 0
Any value multiplied by zero equals zero. Therefore, the mass of the warm water initially in the red cup could be any value, as long as the resulting heat lost is zero.