# Homework Help: Continuous Flow Calorimeter (SHC)

1. Jul 2, 2007

### Lavace

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
VxI = MC(x2 - x1) + q

3. The attempt at a solution
+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.

2. Jul 2, 2007

### Astronuc

Staff Emeritus
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?