Continuous Flow Calorimeter (SHC)

[Lavace]

Homework Statement

380g of a liquid at 12'C in a copper calorimeter weighing 90g is heating at a rate of 20 watt for exactly 3 minutes to produce a temperature of 17'C. If the specific heat capacity of copper is 40Jkg-1K-1, the thermal capacity of the heater is negligible, and there is a negligible heat loss to the surroudnings, obtain a value for the specific heat capacity of the liquid.

Homework Equations

VxI = MC(x2 - x1) + q

The Attempt at a Solution

I genuienly don't have a clue about this.
+q is negligible.

So what are we meant to do for this? I assume we get two values for the equation of the calorimeter and the liquid, minus them and make it C =.
Is that correct? But what values do we sub in for where (I know what everything stands for =/).

Any help is apprectiated!
Thank you.

 

[Astronuc]

Negligible heat is lost to the environment, so the energy stays in the calorimeter and the heat capacity of heater is neglible, which leaves the water and copper.

20 Watts * 180 s = ? Joules (total thermal energy), since 1 W = 1 J/s.

The energy is partitioned between the mass of liquid and mass of copper.

One knows the temperature change, so what is the energy in the copper?

Then knowing that what is the energy in the liquid?

Knowing that - what is the specific heat of liquid?

 

[m2003]

I would approach this problem by first identifying the known values and determining the unknown value, which is the specific heat capacity of the liquid. From the given information, we know the mass of the liquid (380g), the initial and final temperatures (12°C and 17°C), and the heat input (20 watts for 3 minutes). We can also assume that the heat loss to the surroundings and the thermal capacity of the heater are negligible.

To solve for the specific heat capacity of the liquid, we can use the equation VxI = MC(x2 - x1) + q, where V is the voltage, I is the current, M is the mass, C is the specific heat capacity, x2 and x1 are the final and initial temperatures, and q is the heat input.

Since the thermal capacity of the heater and heat loss to the surroundings are negligible, we can simplify the equation to VxI = MC(x2 - x1). We can also rearrange the equation to solve for C, which gives us C = VxI / (M(x2 - x1)).

Substituting the known values, we get C = (20 watts x 3 minutes) / (380g x (17°C - 12°C)) = 0.05 J/g°C. Therefore, the specific heat capacity of the liquid is 0.05 J/g°C.

In conclusion, using the continuous flow calorimeter and the given information, we have determined the specific heat capacity of the liquid to be 0.05 J/g°C. This value can be used in further experiments and calculations involving this liquid.