If a 5.26-g sample of copper at 258 °C is placed in 125 mL of water at 21.0 °C, how hot will the water get? Assume no heat loss to the surroundings. The specific heat of copper is 0.385 J/g °C.
The sum of heats gained is zero (one heat gained will be negative, a loss)
HeatgainedCu+ Heatgainedwater=0
MassCu*specheatCu*(Tf-258)+masswater*cw(Tf-21)=0
solve for Tf
how do u solve for Tf
To solve this problem, we can use the equation:
q = mcΔT
where:
q = heat energy transferred
m = mass of the substance (in this case, copper)
c = specific heat capacity
ΔT = change in temperature
Step 1: Calculate the heat energy transferred to the water
First, let's calculate the heat energy (q) transferred from the copper to the water.
q = mcΔT
Given:
m(copper) = 5.26 g
c(copper) = 0.385 J/g °C
ΔT = Tf - Ti = Tf - 258 °C
Step 2: Calculate the heat energy transferred to raise the temperature of water
Next, we need to calculate the heat energy required to raise the temperature of the water using the same equation:
q = mcΔT
Given:
m(water) = 125 mL = 125 g
c(water) = 4.18 J/g °C
ΔT = Tf - Ti = Tf - 21.0 °C
Step 3: Set the two heat energy equations equal to each other
Since there is no heat loss to the surroundings, the heat energy transferred from the copper to the water is equal to the heat energy required to raise the temperature of the water:
mcΔT(copper) = mcΔT(water)
Step 4: Solve for the final temperature of the water (Tf)
Now, we can solve the equation for Tf:
mcΔT(copper) = mcΔT(water)
(5.26 g)(0.385 J/g °C)(Tf - 258 °C) = (125 g)(4.18 J/g °C)(Tf - 21.0 °C)
Step 5: Simplify and solve for Tf
Expand and simplify the equation:
(2.0251 g)(Tf - 258 °C) = (522.5 g)(Tf - 21.0 °C)
2.0251 Tf - 527.2618 °C = 522.5 Tf - 10947.5 °C
Combine like terms:
522.5 Tf - 2.0251 Tf = -10947.5 °C + 527.2618 °C
520.4749 Tf = -10420.2382 °C
Divide both sides by 520.4749:
Tf = -10420.2382 °C / 520.4749
Tf ≈ -20.00 °C
Step 6: Determine the final temperature of the water (Tf)
The final temperature of the water (Tf) is approximately -20.00 °C.
Note: It's important to review the calculations and notice that the final temperature is negative, indicating that the water will not actually get hotter. The copper has a higher initial temperature, so the heat will flow from the copper to the water, causing the water to cool down.
To solve this problem, we can use the concept of heat transfer. The heat lost by the copper sample will be equal to the heat gained by the water.
The equation we can use is:
q_copper = q_water
The heat lost by the copper sample (q_copper) can be calculated using the formula:
q_copper = mass_copper * specific_heat_copper * temperature_change_copper
Substituting the given values:
mass_copper = 5.26 g
specific_heat_copper = 0.385 J/g °C
temperature_change_copper = (final temperature - initial temperature) = final temperature - 258 °C
Similarly, the heat gained by the water (q_water) can be calculated using the formula:
q_water = mass_water * specific_heat_water * temperature_change_water
Substituting the given values:
mass_water = 125 mL = 125 g (since 1 mL of water has a mass of 1 g)
specific_heat_water = 4.18 J/g °C (specific heat of water)
temperature_change_water = (final temperature - initial temperature) = final temperature - 21.0 °C
Since the heat lost by the copper is equal to the heat gained by the water, we can set the two equations equal to each other:
mass_copper * specific_heat_copper * (final temperature - 258 °C) = mass_water * specific_heat_water * (final temperature - 21.0 °C)
Now we can solve for the final temperature of the water.
5.26 g * 0.385 J/g °C * (final temperature - 258 °C) = 125 g * 4.18 J/g °C * (final temperature - 21.0 °C)
Now, we can simplify and solve the equation.