50 gram of steel at 100 degree centigrade is addae to 100grame of water at 30 degree centigrade the resultant temperature is 35 degree centigrade find the specific heat of metel
heat lost by metal + heat gained by H2O = 0
[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute the numbers and solve for specific heat metal which is the only unknown in th equation. BTW, centigrade is out; celsius is in.
To find the specific heat of the metal, we can use the formula:
Q = m * c * ΔT
Where:
Q = heat gained or lost by the substance
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
In this case, the heat gained by the steel is equal to the heat lost by the water. Let's calculate the heat gained by the steel and the heat lost by the water separately.
Heat gained by the steel:
Q1 = m1 * c1 * ΔT1
m1 = 50 grams (mass of steel)
c1 = specific heat capacity of steel (to be found)
ΔT1 = change in temperature for steel = (resultant temperature - initial temperature of steel) = (35°C - 100°C) = -65°C
Heat lost by the water:
Q2 = m2 * c2 * ΔT2
m2 = 100 grams (mass of water)
c2 = specific heat capacity of water = 4.186 J/g°C (assuming it's water at room temperature)
ΔT2 = change in temperature for water = (resultant temperature - initial temperature of water) = (35°C - 30°C) = 5°C
Since the heat gained by the steel is equal to the heat lost by the water, we can set Q1 equal to Q2:
Q1 = Q2
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Substituting the values we know:
50 * c1 * (-65) = 100 * 4.186 * 5
Simplifying:
-3250 * c1 = 2093
Finally, we can solve for c1:
c1 = 2093 / -3250
c1 ≈ -0.643 J/g°C
Therefore, the specific heat capacity of the metal is approximately -0.643 J/g°C.
To find the specific heat of the metal, we can use the principle of heat transfer, which states that the heat gained by the metal is equal to the heat lost by the water.
The formula for heat transfer is Q = mcΔT, where:
- Q is the heat gained or lost
- m is the mass of the substance
- c is the specific heat
- ΔT is the change in temperature
In this scenario:
- The mass of the steel is 50 grams.
- The initial temperature of the steel is 100 degrees Celsius.
- The mass of the water is 100 grams.
- The initial temperature of the water is 30 degrees Celsius.
- The final temperature of both substances is 35 degrees Celsius.
Using the formula mentioned above, we can set up two equations, one for each substance:
For the steel:
Q1 = mcΔT1
For the water:
Q2 = mcΔT2
Since the system is isolated and the heat lost by the steel is equal to the heat gained by the water, we can equate the two equations:
Q1 = Q2
mcΔT1 = mcΔT2
Since the specific heat of the water (c) is known (4.186 J/g°C), we can rewrite the equation as:
m1c1ΔT1 = m2c2ΔT2
Plugging in the known values:
(50g)(c1)(35 - 100) = (100g)(4.186 J/g°C)(35 - 30)
To find the specific heat of the metal (c1), we can solve this equation.