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

To calculate the energy lost by the hot copper, we use the formula:

$E = m imes c imes riangle T$

where:

• $m$ is the mass of the copper: $m = 101.2 ext{ g} - 55.7 ext{ g} = 45.5 ext{ g} = 0.0455 ext{ kg}$
• $c$ is the specific heat capacity of copper: $c = 390 ext{ J kg}^{-1} ext{ K}^{-1}$
• $riangle T$ is the change in temperature: $riangle T = 99.5 ext{ °C} - 21.0 ext{ °C} = 78.5 ext{ °C}$

Plugging in the values:

$E = 0.0455 ext{ kg} imes 390 ext{ J kg}^{-1} ext{ K}^{-1} imes 78.5 ext{ K}$

Calculating gives:

$E ext{ (energy lost)} = 9247.57 ext{ J}$

To find the specific heat capacity of water, we first calculate the heat gained by the water:

$ext{Heat gained by water} = ext{mass of water} imes c_w imes ext{temperature rise}$

The mass of water can be calculated as:

$ext{mass of water} = 101.2 ext{ g} - 55.7 ext{ g} = 45.5 ext{ g} = 0.0455 ext{ kg}$

Using the final and initial temperatures, the temperature rise is:

$ext{Temperature rise} = 21.0 ext{ °C} - 16.5 ext{ °C} = 4.5 ext{ °C}$

Now we set up the equation:

$ext{Heat lost by copper} = ext{Heat gained by water}$

Plugging in the values gives:

$9245.7 ext{ J} = 0.0455 ext{ kg} imes c_w imes 4.5 ext{ K}$

Solving for $c_w$, we find:

c_w = rac{9245.7 ext{ J}}{0.0455 ext{ kg} imes 4.5 ext{ K}} = 4038 ext{ J kg}^{-1} ext{ K}^{-1}

1. The calorimeter was insulated to reduce heat transfer with the environment, using materials that prevent heat loss.
2. The experiment was conducted quickly to minimize exposure time, ensuring that the water and copper did not lose heat to the surroundings during the measurements.

Adding a larger mass of copper would reduce the percentage error because it would minimize the effect of heat loss during the transfer process. A larger mass retains heat better, leading to a smaller error in temperature change, thus improving the accuracy of the specific heat capacity calculation.