Calculate the heat capacity of this calorimeter

[dg_5021]

When using a different calorimeter, and mixing 50 ml of hot water at 65 C with 60 ml of water in the calorimeter at 25 C, the temperture of the calorimeter increased by 5.5 C. Calculate the heat capacity of this calorimeter. Is this a better or worse calorimeter than the one used in the sample calculations?

This problem gets me all confused? Is there a formula that would make it easier? please help

 

[so-crates]

As with all problems, if they ask for X. The first thing you want to ask yourself is, what is the definition of X? In this case heat capacity.

 

[ravenprp]

Yes, there is a formula that can help you solve this problem. The heat capacity of a system is defined as the amount of heat required to raise the temperature of the system by 1 degree Celsius. In this case, the system is the calorimeter and the heat capacity is the unknown value we are trying to find.

The formula for heat capacity is: C = Q/ΔT, where C is the heat capacity, Q is the amount of heat transferred, and ΔT is the change in temperature.

In the given scenario, we know that 50 ml of hot water at 65 C is mixed with 60 ml of water in the calorimeter at 25 C, resulting in a temperature increase of 5.5 C. This means that 5.5 C is the change in temperature (ΔT) and the amount of heat transferred (Q) is the same for both the hot water and the calorimeter-water mixture.

We can calculate the amount of heat transferred using the formula Q = m × c × ΔT, where m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. Since the mass of the hot water and the calorimeter-water mixture are the same, we can set their respective heat transfer values equal to each other.

m1 × c1 × ΔT = m2 × c2 × ΔT

Since ΔT is the same for both substances, we can cancel it out, leaving us with:

m1 × c1 = m2 × c2

We know that the specific heat capacity of water (c) is 4.186 J/g°C. Plugging in the values for the mass and specific heat capacity of the hot water (50 ml and 4.186 J/g°C, respectively) and the calorimeter-water mixture (60 ml and unknown heat capacity, C), we get:

50 ml × 4.186 J/g°C = 60 ml × C

Solving for C, we get:

C = (50 ml × 4.186 J/g°C) / 60 ml

C = 3.490 J/g°C

Therefore, the heat capacity of this calorimeter is 3.490 J/g°C.

To determine if this is a better or worse calorimeter than the one used in the sample calculations, we would need to know the heat capacity of the previous calorimeter. However, based on the