Question

A beaker containing 250.0 g of water is heated with 1500.0 J of heat energy. If the temperature of the water changed from 22.0000oC to 23.4354oC, what is the specific heat of water?

Answer 1

Answer on the question #70258, Chemistry / Other

Question:

A beaker containing 250.0 g of water is heated with 1500.0 J of heat energy. If the temperature of the water changed from 22.0000°C to 23.4354°C, what is the specific heat of water?

Answer:

The relation between the heat transferred to body and change in its temperature is the following:

$$Q = c m \Delta T,$$

where $c$ is the specific heat of the body, $m$ is its mass and $\Delta T$ is the change in its temperature.

Thus, we can easily find the specific heat:

c = \frac{Q}{m \Delta T} = \frac{1500.0 \, \text{J}}{250.0 \, g \cdot (23.4354 - 22.0000)^\circ \text{C}} = 4.180 \, \text{J} \, \text{g}^{-1} \, ^\circ \text{C}^{-1}

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