A piece of metal of mass 112g is heated 100 degree C.,and dropped into a copper calorimeter of mass 40g containing 200g of water at 16 derreC. Neglecting the heat loss, the specific het of the meatal is nearly,if the equilibrium temperature reached is 24.1degree C,
OPTIONS :
1.0.194 cal?gm degree C.
2.0.294 cal?gm degree C.
m1•c1•(100-24.1)= m2•c2•(24.1-16) + m3•v3•(24.1-16),
c1 = (m2•c2+ m3•v3)•8.1/m1•75.9 =
= 811 J/kg•degr =0.194 cal/g•degr
To find the specific heat of the metal, we can use the heat transfer 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, we need to find the heat transferred to the water in the calorimeter:
Q_water = mcΔT
Q_water = (200g)(1 cal/g°C)(24.1°C - 16°C)
Q_water = 1600 cal
Next, we need to find the heat transferred from the metal to the water:
Q_metal = mcΔT
Q_metal = (112g)(c)(24.1°C - 100°C)
Since the system is in thermal equilibrium, the heat transferred by the metal is equal to the heat absorbed by the water:
Q_metal = Q_water
(112g)(c)(24.1°C - 100°C) = 1600 cal
Now, let's solve for c:
c = 1600 cal / ((112g)(-75.9°C))
Simplifying the equation:
c ≈ 0.194 cal/g°C
Therefore, the specific heat of the metal is approximately 0.194 cal/g°C.
So, the correct option is 1. 0.194 cal/g°C.
To solve this problem, we can use the principle of heat transfer which states that the heat gained by the water and the calorimeter is equal to the heat lost by the metal.
First, we need to calculate the heat gained by the water and the calorimeter. We can use the formula:
Q = mcΔT
where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the water and the calorimeter, we have:
Qwc = (mwater + mcalorimeter)cwaterΔTwc
Next, we calculate the heat lost by the metal. We can use the formula:
Qm = mcΔT
where Qm is the heat lost by the metal, m is the mass, c is the specific heat capacity of the metal, and ΔT is the change in temperature.
For the metal, we have:
Qm = mmeltingcmetalΔTm
Since the heat gained by the water and the calorimeter is equal to the heat lost by the metal, we have:
Qwc = Qm
Substituting the values, we get:
(mwater + mcalorimeter)cwaterΔTwc = mmeltingcmetalΔTm
We are given the following values:
mwater = 200g
mcalorimeter = 40g
cwater = 1 cal/g°C (specific heat capacity of water)
ΔTwc = 24.1°C - 16°C = 8.1°C
ΔTm = 24.1°C - 100°C = -75.9°C (since the metal is losing heat)
Substituting these values into the equation, we can solve for cmetal.
(200g + 40g)(1 cal/g°C)(8.1°C) = 112g * cmetal * (-75.9°C)
240 * 8.1 = 112 * cmetal * (-75.9)
After performing the calculations, we find that cmetal ≈ 0.294 cal/g°C.
Therefore, the specific heat of the metal is approximately 0.294 cal/g°C.
So, the correct option is 2. 0.294 cal/g°C.