A 50g chuck of metal (c=0.506 j/g degrees celsius) is heated to 60 degrees celsius. It is then dropped into a calorimeter (c=1.40 j/g derees celsius, m - 5.0g) that contains 100g of liquid. The temperature of the liquid raises from 15 degrees celsius to 20 degrees celsius. What is the celsius of the liquid?
To solve this problem, we can use the principle of conservation of energy. The heat gained by the liquid is equal to the heat lost by the metal. We can calculate the heat gained by the liquid using the equation:
Q = m * c * ΔT
Where:
Q = heat gained (or lost)
m = mass of the liquid (in grams)
c = specific heat capacity of the liquid (in J/g °C)
ΔT = change in temperature (final temperature - initial temperature)
In this case, the mass of the liquid is 100g, the specific heat capacity of the liquid is 1.40 J/g °C, and the change in temperature is 20°C - 15°C = 5°C. Therefore:
Q = 100g * 1.40 J/g °C * 5°C
Q = 700 J
Now, we can equate the heat gained by the liquid to the heat lost by the metal, using the same equation:
Q = m * c * ΔT
The mass of the metal is 50g, and the specific heat capacity of the metal is 0.506 J/g °C. The initial temperature of the metal is 60°C and the final temperature is the same as the liquid, which is what we are trying to find. Let's call it T:
700 J = 50g * 0.506 J/g °C * (60°C - T)
Now we can solve for T:
700 J = 30.3 J °C/g * (60°C - T)
700 J = 1818 J °C - 30.3 J °C/g * T
1818 J °C - 700 J = 30.3 J °C/g * T
1118 J °C = 30.3 J °C/g * T
T = 1118 J °C / 30.3 J °C/g
T ≈ 36.86°C
Therefore, the final temperature of the liquid is approximately 36.86°C.