Question

the water absorbs about 2120 j and the calorimeter absorbs about 22 j. what is the heat change for the unknown material?
$q_{unknown} = ? j$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Understand heat transfer

In a calorimetry setup, the heat lost by the unknown material is equal to the heat gained by the water and the calorimeter (assuming no heat loss to the surroundings). So, \( q_{\text{unknown}} = - (q_{\text{water}} + q_{\text{calorimeter}}) \).

Step2: Calculate total heat gained

First, find the total heat gained by water and calorimeter: \( q_{\text{water}} + q_{\text{calorimeter}} = 2120\ \text{J} + 22\ \text{J} = 2142\ \text{J} \).

Step3: Determine heat change for unknown

Since the unknown material loses heat (the water and calorimeter gain it), the heat change for the unknown is the negative of the total heat gained: \( q_{\text{unknown}} = - 2142\ \text{J} \).

Answer:

\(-2142\)