Question

A piece of metal was heated and then put it into 100.0 ml of water, initially at 21.2 degrees celsius. The metal and water were allowed to come to an equiLiBrium temperature, determined to be 32.0 degrees celsius. How much heat did the water absorb?

Answer 1

To determine the heat absorbed by water, use q = mcΔT with the specific heat capacity of water (4.18 J/g°C), the mass of water (equivalent to its volume in mL for water), and the actual change in temperature.

Step-by-step explanation:

To calculate how much heat the water absorbed, you need to use the formula q = mcΔT, where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. In this case, the specific heat capacity (c) of water is typically 4.18 J/g°C. To find m, the mass of water, we use the density of water which is approximately 1 g/mL to convert 100.0 mL to 100.0 g (since the density of water is 1 g/mL, the mass is numerically equivalent to the volume in mL).

The change in temperature (ΔT) is the final temperature minus the initial temperature, which is 32.0 °C - 21.2 °C = 10.8 °C. Using the formula: q = (100.0 g)(4.18 J/g°C)(10.8 °C), we can find the heat absorbed by the water.