A 355-g piece of metal at 48.0 degree celsius is dropped into 111g of water at 15.0 degree celsius.If the final temperature of the mixture is 32.5 degree celsius what is the specific heat of the metal?
Q1 =Q2
c1•m1 •Δt1 = c2•m2• Δt2
c1 = c2•m2• Δt2/ m1 •Δt1 =
=4180•0.111•17.5/0.335•15.5 = 1564 J/kg•oC
To find the specific heat of the metal, we can use the equation:
q = mcΔT
where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
First, let's calculate the heat transferred from the metal to the water:
q1 = mcΔT
where m is the mass of the metal, c is the specific heat of the metal, and ΔT is the change in temperature.
Given that the mass of the metal is 355 g and the change in temperature is (32.5 - 48.0) °C = -15.5 °C, we can calculate q1:
q1 = (355 g)(c)(-15.5 °C)
Next, let's calculate the heat transferred from the water to the metal:
q2 = mcΔT
where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature.
Given that the mass of the water is 111 g and the change in temperature is (32.5 - 15.0) °C = 17.5 °C, we can calculate q2:
q2 = (111 g)(4.18 J/g°C)(17.5 °C)
Since the heat transferred from the metal to the water is equal to the heat transferred from the water to the metal, we can set q1 equal to q2:
(355 g)(c)(-15.5 °C) = (111 g)(4.18 J/g°C)(17.5 °C)
Simplifying and solving for c:
355c(-15.5) = (111)(4.18)(17.5)
-5492.5c = 8213.925
c = 8213.925 / -5492.5
c = -1.495 J/g°C
Therefore, the specific heat of the metal is approximately -1.495 J/g°C.
To find the specific heat of the metal, you can use the principle of conservation of energy. The energy gained or lost by the metal and water must be equal.
First, let's calculate the heat lost by the metal:
1. Calculate the heat lost by the metal using the formula:
Q = m × c × ΔT
where Q is the heat lost, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
- Mass of the metal (m) = 355 g
- Initial temperature of the metal (T1) = 48.0°C
- Final temperature of the mixture (Tf) = 32.5°C
ΔT = Tf - T1
= 32.5°C - 48.0°C
= -15.5°C (since the metal lost heat)
Q = 355 g × c × -15.5°C
Next, let's calculate the heat gained by the water:
2. Calculate the heat gained by the water using the formula:
Q = m × c × ΔT
where Q is the heat gained, m is the mass, c is the specific heat, and ΔT is the change in temperature.
Given:
- Mass of the water (m) = 111 g
- Initial temperature of the water (T1) = 15.0°C
- Final temperature of the mixture (Tf) = 32.5°C
ΔT = Tf - T1
= 32.5°C - 15.0°C
= 17.5°C (since the water gained heat)
Q = 111 g × c × 17.5°C
Now, since the energy gained by the water is equal to the energy lost by the metal (law of conservation of energy), we can set up the equation:
Q_lost by metal = Q_gained by water
355 g × c × -15.5°C = 111 g × c × 17.5°C
Simplifying the equation:
-15.5°C/c = 17.5°C/c
Now, cross-multiply and solve for c:
-15.5°C × c = 17.5°C × c
-15.5 = 17.5
This equation implies that there was an error in the calculation because it's not possible for -15.5 to be equal to 17.5. Please double-check the given values and re-calculate.