Question

95.0 g of an unknown metal at 100°C is placed in 800 g of water at 25.0°C. The final temperature of the water is 29.8°C. What was the specific heat of the metal? • The specific heat of water is: Cho = 4.184 Select the correct answer below: 2.41 O 1.89 O 166 O 1.25

Answer 1

To find the specific heat of the metal, we use the equation q_metal = m_metal * c_metal * ΔT. Setting the equation equal to the equation for the heat gained or lost by the water, we can solve for the specific heat of the metal. Plugging in the given values, the specific heat of the metal is approximately 1.89 J/g°C.

Step-by-step explanation:

To find the specific heat of the metal, we can use the equation:

Qmetal = Mmetal * Cmetal * ΔT

Qmetal is the heat gained or lost by the metal

Mmetal is the mass of the metal

Cmetal is the specific heat of the metal

ΔT is the change in temperature of the metal

Similarly, we can use the equation:

Qwater = Mwater * Cwater * ΔT

Qwater is the heat gained or lost by the water

Mwater is the mass of the water

Cwater is the specific heat of water

ΔT is the change in temperature of the water

Since the final temperature of the water is the same as the final temperature of the metal, we can set the two equations equal to each other:

Mmetal * Cmetal * ΔT = Mwater * Cwater * ΔT

Simplifying the equation gives us:

Cmetal = (Mwater * Cwater * ΔT) / Mmetal

Plugging in the given values, we get:

Cmetal = (800 g * 4.184 J/g°C * (29.8°C - 25.0°C)) / 95.0 g

Solving the expression, we find that the specific heat of the metal is approximately 1.89 J/g°C