The specific heat capacity of water was found by adding hot copper to water in a copper calorimeter - Leaving Cert Physics - Question 2 - 2007

The energy lost by the hot copper can be calculated using the formula:

$E = m imes c imes riangle T$

Where:

• $m = 0.32 imes 10^{3} ext{g} = 320 ext{g}$
• $c = 390 ext{J kg}^{-1} ext{K}^{-1}$
• $riangle T = (99.5°C - 21.0°C) = 78.5°C$

Converting grams to kilograms:

• $m = 0.32 ext{ kg}$

Thus, the energy can be calculated as:

$E = 0.32 imes 10^{3} imes 390 imes 78.5 = 9245.57 ext{ J}$

or approximately 9246 J.

The specific heat capacity of water can be calculated from the heat gained by water, which is equal to the heat lost by the copper:

$9247.50 = (mass ext{ of water}) imes (c_{w}) imes (final ext{ temp} - initial ext{ temp})$

Taking the mass of water as $m_{w} = 101.2 ext{g} - 55.7 ext{g} = 45.5 ext{g} = 0.0455 ext{ kg}$ and substituting into the equation gives:

$c_{w} = \frac{E}{m_{w} imes (T_{final} - T_{initial})}$

Substituting in the values:

$c_{w} = \frac{9246}{0.0455 imes (21.0 - 16.5)} = 4.038 imes 10^{3} ext{ J kg}^{-1} ext{ K}^{-1}$

Thus, the specific heat capacity of water is approximately $4.04 imes 10^{3} ext{ J kg}^{-1} ext{ K}^{-1}$.

1. Insulating the calorimeter with insulating materials (like a lid or insulating foam) helps reduce heat loss.
2. Avoiding splashes and ensuring that the calorimeter is not exposed to air will help maintain temperature.

Adding a larger mass of copper would ensure a larger heat transfer, resulting in a more noticeable temperature change in the water. This improves the sensitivity of the measurements, as it minimizes the effects of heat loss and provides more data for calculating the specific heat capacity.