A piece of metal with a mass of 1.9 kg, specific
heat of 200 J/kg
◦
C, and initial temperature
of 100
◦
C is dropped into an insulated jar that
contains liquid with a mass of 3.9 kg, specific
heat of 1000 J/kg
◦
C, and initial temperature
of 0
◦
C. The piece of metal is removed after
7 s, at which time its temperature is 20
◦
C.
Find the temperature of the liquid after the
metal is removed. Neglect any effects of heat
transfer to the air or to the insulated jar.
Answer in units of
◦
C
the sum of the heats gained is zero.
massmetal*cmetal(20-100)+massliquid*cliquid(Tf-0)=0
solve for Tf
To find the temperature of the liquid after the metal is removed, we can use the principle of conservation of energy.
The energy gained by the metal is equal to the energy lost by the liquid.
The energy gained by the metal can be calculated using the formula:
Q = mcΔT
where Q is the heat energy gained, m is the mass of the metal, c is the specific heat of the metal, and ΔT is the change in temperature of the metal.
Substituting the given values, we have:
Q_metal = (1.9 kg) * (200 J/kg◦C) * (20◦C - 100◦C)
The energy lost by the liquid can be calculated using the formula:
Q = mcΔT
where Q is the heat energy lost, m is the mass of the liquid, c is the specific heat of the liquid, and ΔT is the change in temperature of the liquid.
Substituting the given values, we have:
Q_liquid = (3.9 kg) * (1000 J/kg◦C) * (T - 0◦C)
Since energy is conserved, we can set Q_metal equal to Q_liquid:
Q_metal = Q_liquid
(1.9 kg) * (200 J/kg◦C) * (20◦C - 100◦C) = (3.9 kg) * (1000 J/kg◦C) * (T - 0◦C)
Simplifying the equation:
-15240 J = 3900 kg J/◦C * T
Solving for T:
T = -15240 J / (3900 kg J/◦C)
T ≈ -3.91◦C
Therefore, the temperature of the liquid after the metal is removed is approximately -3.91◦C.