Calorimetry Problem - Find Specific Heat of Liquid

[Jacob G]

If anyone could possibly help me and explain how to do this problem it would be much appreciated..

A 50g copper mass is heated by placing it in boiling water. It is then placed in a beaker containing 250g of an unknown liquid at 20C. The final temperature of the weight and liquid is found to be 25C. What is the specific heat of the liquid? (Assume no heat is lost to the surroundings.)

Thanks..

 

[HallsofIvy]

That looks like a homework problem to me! You know the initial temperature of the copper mass and you know its mass and can look up its specific heat. That allows you to calculate the amount of heat energy in the mass before it was put in the lquid. Knowing that its final temperature was 25 C, you can do the same thing to find the final amount of hear energy in the mass. The difference is the energy transferred to the liquid. Again, you know the initial and final temperatures of the liquid as well as its mass so you can write down the initial and final amounts of heat energy in the liquid. Set their difference equal to the heat energy transferred from the copper to the liquid and solve for the specific heat.

 

[Moetasim]

To find the specific heat of the liquid, we can use the formula Q = mcΔT, where Q is the heat absorbed or released, m is the mass of the substance, c is the specific heat, and ΔT is the change in temperature. In this case, we know the mass of the copper (50g), the final temperature (25C), and the initial temperature of the liquid (20C). We also know that no heat is lost to the surroundings, so the heat absorbed by the copper must be equal to the heat released by the liquid.

First, we can calculate the heat absorbed by the copper using the formula Q = mcΔT. Since the copper is heated from boiling water to the final temperature of 25C, we can use the specific heat of copper (0.385 J/g°C) and the change in temperature (100C) to calculate the heat absorbed by the copper. This gives us a value of 1925 J.

Next, we can equate this to the heat released by the liquid. Using the same formula, Q = mcΔT, we can plug in the mass of the liquid (250g), the specific heat (which we are trying to find), and the change in temperature (5C). This will give us an equation of 1925 J = (250g)(c)(5C).

Solving for c, we get c = 1.54 J/g°C. Therefore, the specific heat of the unknown liquid is 1.54 J/g°C. I hope this helps and let me know if you have any further questions.