Question

A sample of metal that has a mass of 12.48 g is heated to 99.0°C and then dropped into 50.0mL of 25°C water, causing the temperature of the water to become 28.1°C. What is the specific heat of the metal?

Answer 1

To determine the specific heat of the metal and its identity, use the equation q_metal = -q_water.

Step-by-step explanation:

We can use the equation: qmetal = -qwater where qmetal is the heat lost by the metal and qwater is the heat gained by the water. First, calculate the heat gained by the water using the equation qwater = mcΔT, where m is the mass of the water, c is its specific heat capacity, and ΔT is the change in temperature:

qwater = 50.0 mL × 1 g/mL × 4.18 J/g°C × (28.1 - 25.0)°C. Next, calculate the heat lost by the metal using the equation qmetal = mcΔT, where m is the mass of the metal, c is its specific heat capacity, and ΔT is the change in temperature.

Rearrange the equation to solve for c: c = qmetal / (mΔT). Plug in the values and calculate c. The identity of the metal can be determined by comparing the specific heat capacity of the metal to the known specific heat capacities of different metals.