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What is the specific heat capacity of a metal if a 56.8 g block of the metal at 92.2 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)

User Pari
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Final answer:

To find the specific heat capacity of the metal, we use the heat transfer equation and apply the conservation of energy, equating the heat lost by the metal to the heat gained by the water, then solving for the metal's specific heat capacity.

Step-by-step explanation:

To calculate the specific heat capacity of a metal using a calorimetry experiment, we apply the principle of conservation of energy which implies that the heat lost by the hot metal will be equal to the heat gained by the colder water. The equation for the heat transfer is:

q = mcΔT

where q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the water, the heat gained (q water) can be calculated using:

q water = (125 g)(4.184 J/g°C)(24.2 °C - 18.5 °C)

For the metal, the heat lost (q metal) can be calculated assuming it is equal to the heat gained by the water, thus :

q metal = (56.8 g)(c metal)(24.2 °C - 92.2 °C) = -q water

By solving these equations, we can find the specific heat capacity of the metal (c metal). The final temperature of both water and the metal is the same (24.2 °C), which represents the thermal equilibrium state.

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