Question

A calorimeter contains 1.0 kg of water (specific heat = 4,184 J/(kg.°C)). An object with a mass of 4.23 kg is added to the water. If the water temperature increases by 3.0°C and the temperature of theobject decreases by 1.0°C, what is the specific heat of the object?

Answer 1

To find the specific heat of the object, we can use the principle of energy conservation. The energy gained by the water is equal to the energy lost by the object. Using Q = mcΔT, we can calculate the specific heat of the object as 2962.3 J/(kg.°C).

Step-by-step explanation:

To find the specific heat of the object, we can use the principle of energy conservation. The energy gained by the water is equal to the energy lost by the object. We can calculate the energy gained by the water using the formula: Q = mcΔT, where Q is the heat gained, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the given values, we have: Q = (1.0 kg)(4184 J/(kg.°C))(3.0°C) = 12,552 J. Since the energy lost by the object is equal to the energy gained by the water, we have 12,552 J = (4.23 kg)(c)(-1.0°C). Solving for c, we find that the specific heat of the object is 2962.3 J/(kg.°C).