if 50 grams of a substance with temperature of 20 degrees C and a specific heat of 2 is mixed with 200 grams of water at 40 degrees C, what will be the equilibrium temperature of the mixture?

heat gained by first substance + heat lost by water = 0

[mass*specific heat x (Tfinal-Tinitial)] + [massH2O x specific heat water x (Tfinal-Tnitial)] = 0
Solve for Tfinal.
Post your work if you get stuck.

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To calculate the equilibrium temperature of the mixture, we can use the principle of conservation of energy, also known as the heat equation. The heat equation states that:

Q = m * c * ΔT

where:
Q is the heat exchanged,
m is the mass of the substance,
c is the specific heat, and
ΔT is the change in temperature.

In this case, we have two substances with different initial temperatures and masses. Let's calculate the heat exchanged by both substances separately:

1. Heat exchanged by the substance:
Qsubstance = msubstance * csubstance * ΔTsubstance

Given that:
msubstance = 50 grams
csubstance = 2
ΔTsubstance = (equilibrium temperature - initial temperature of the substance)

Since the initial temperature of the substance is 20°C, we have:
ΔTsubstance = (equilibrium temperature - 20)

2. Heat exchanged by water:
Qwater = mwater * cwater * ΔTwater

Given that:
mwater = 200 grams
cwater = 1 (specific heat of water)
ΔTwater = (equilibrium temperature - initial temperature of water)

Since the initial temperature of the water is 40°C, we have:
ΔTwater = (equilibrium temperature - 40)

Now, according to the principle of conservation of energy, the heat gained by one substance must be equal to the heat lost by the other substance when they reach equilibrium. Therefore:

Qsubstance = -Qwater

By substituting the above equations, we get:
msubstance * csubstance * ΔTsubstance = -mwater * cwater * ΔTwater

Plugging in the given values and rearranging the equation, we have:
50 * 2 * (equilibrium temperature - 20) = -200 * 1 * (equilibrium temperature - 40)

Now, we can solve for the equilibrium temperature:

100 * (equilibrium temperature - 20) = -200 * (equilibrium temperature - 40)

100 * equilibrium temperature - 2000 = -200 * equilibrium temperature + 8000

Combining like terms:
300 * equilibrium temperature = 10000

Solving for equilibrium temperature:
equilibrium temperature = 10000 / 300
equilibrium temperature ≈ 33.33°C

Therefore, the equilibrium temperature of the mixture will be approximately 33.33°C.