when 45g of an alloy is dropped into 100.0 g of water at 25c, the final temperature is 37c. what is the specific heat of the alloy
heat lost by alloy + heat gained by water = 0
[mass alloy x specific heat alloy x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
To find the specific heat of the alloy, we can use the formula:
heat gained by water = heat lost by alloy
The heat gained or lost can be determined using the formula:
Q = mcΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, the heat gained by the water is equal to the heat lost by the alloy.
Given:
Mass of water (m₁) = 100.0 g
Initial temperature of water (T₁) = 25°C
Final temperature of water (T₂) = 37°C
Using the formula for the water:
Q(water) = m₁c(water)ΔT(water)
Now, let's find the heat lost by the alloy.
Mass of the alloy (m₂) = 45g
Specific heat of the alloy (c_alloy) = ?
Using the formula for the alloy:
Q(alloy) = m₂c_alloyΔT(alloy)
Since the heat lost by the alloy is equal to the heat gained by the water:
Q(water) = Q(alloy)
m₁c(water)ΔT(water) = m₂c_alloyΔT(alloy)
Now, we can substitute the given values into the equation and solve for the specific heat of the alloy (c_alloy).
m₁c(water)(T₂ - T₁) = m₂c_alloy(T₂ - T₁)
(100.0 g)(4.18 J/g°C)(37°C - 25°C) = (45 g)(c_alloy)(37°C - 25°C)
Simplifying:
(100.0 g)(12°C) = (45 g)(c_alloy)(12°C)
1200 = 540c_alloy
Divide both sides by 540:
1200 / 540 = c_alloy
c_alloy ≈ 2.22 J/g°C
Therefore, the specific heat of the alloy is approximately 2.22 J/g°C.
To find the specific heat of the alloy, we can use the equation:
m1c1ΔT1 = m2c2ΔT2
where:
m1 = mass of the alloy (45 g)
c1 = specific heat of the alloy (to be determined)
ΔT1 = change in temperature of the alloy (final temperature - initial temperature)
m2 = mass of water (100.0 g)
c2 = specific heat of water (4.18 J/g°C)
ΔT2 = change in temperature of water (final temperature - initial temperature)
Let's plug in the given values into the equation and solve for c1:
(45 g)(c1)(37°C - 25°C) = (100.0 g)(4.18 J/g°C)(37°C - 25°C)
12(c1) = 1000(4.18)
12(c1) = 4180
c1 = 4180 J/(12 g x 8°C)
c1 = 43.75 J/(g°C)
Therefore, the specific heat of the alloy is 43.75 J/(g°C).