Question

To measure the heat capacity of an object, all you usually have to do is put it in thermal contact with another object whose heat capacity is known. Suppose that a 100 g chunk of metal is immersed in boiling water (100 oC). After a time, the metal is removed and quickly transferred to a Styrofoam cup containing 250 g of water at (280C). After a little while, the temperature of the contents of the cup is found to be 44 oC. You may assume that the heat transferred from contents of the cup to its surroundings, and the heat capacity of the cup itself are both negligible.

(a) How much heat is gained by the water in the Styrofoam cup?
(b) How much heat is lost by the metal?
(c) What is the heat capacity of the metal?
(d) What is the specific heat (in J/g K) of the metal?

Answer 1

The heat gained by the water in the Styrofoam cup is 3,370 J. The heat lost by the metal is also 3,370 J. The heat capacity and specific heat of the metal are both 3.37 J/g°C.

Step-by-step explanation:

To calculate the heat gained by the water in the Styrofoam cup, we can use 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 values, we get Q = (250 g)(4.18 J/g°C)(44°C - 28°C) = 3,370 J.

To calculate the heat lost by the metal, we can use the same formula. The mass of the metal is given as 100 g, and assuming its specific heat capacity is unknown, we'll use the symbol Cp for now. We can write Q = (100 g)(Cp)(100°C - 44°C).

Since the heat gained by the water is equal to the heat lost by the metal (as per the law of conservation of energy), we can set the two equations equal to each other: 3,370 J = (100 g)(Cp)(100°C - 44°C). Solving for Cp, we find Cp = 3.37 J/g°C. Therefore, the heat capacity of the metal is 3.37 J/g°C, and its specific heat is also 3.37 J/g°C.