Question

What is the specific heat capacity of a metal if a 65.9 g block of the metal at 90.7 oc causes 125 g of water at 18.5 oc to warm to 24.2 oc (assume metal's final temperature is also 24.2 oc)? (specific heat capacity of water is 4.184 j/g*oc)

Answer 1

The specific heat capacity of a metal can be found using a calorimetry experiment with the equation q = m x c x ΔT, where m is mass, c is specific heat capacity, and ΔT is the change in temperature. By setting the heat gained by water equal to the heat lost by the metal, one can solve for the metal's specific heat.

Step-by-step explanation:

Calculating the Specific Heat Capacity of a Metal

To determine the specific heat capacity of a metal in a calorimetry experiment, one can use the principle of conservation of energy which states that the heat lost by the hot metal will be equal to the heat gained by the colder water. The formula that connects the mass, temperature change, and specific heat capacity is q = m x c x ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat lost or gained can be calculated using the relationship:

q = m x c x ΔT

For the water, we have:

q_water = (mass of water) x (specific heat of water) x (ΔT of water)

For the metal, we have:

q_metal = (mass of metal) x (specific heat of metal) x (ΔT of metal)

Since the water heats up and the metal cools down:

|q_metal| = |q_water|

We can then solve for the specific heat of the metal by rearranging the terms.