Question

A piece of metal with a mass of 15 grams initially at 25°C is placed in 50 mL of water at 80°C. After the metal has reached thermal equilibrium with the water, its temperature is 30°C. Calculate the specific heat capacity of the metal.

Answer 1

To calculate the specific heat capacity of the metal, we can use the formula:Q = mcTwhere Q is the heat transferred, m is the mass, c is the specific heat capacity, and T is the change in temperature.When the metal reaches thermal equilibrium with the water, the heat lost by the water is equal to the heat gained by the metal. Therefore, we can set up the following equation:Q_water = Q_metalFor water, we know the specific heat capacity  c_water  is 4.18 J/gC, and the mass of water is 50 grams  since 1 mL of water has a mass of approximately 1 gram . The change in temperature for water  T_water  is the initial temperature  80C  minus the final temperature  30C , which is 50C.For the metal, we have the mass  m_metal  as 15 grams, and we need to find the specific heat capacity  c_metal . The change in temperature for the metal  T_metal  is the final temperature  30C  minus the initial temperature  25C , which is 5C.Now we can set up the equation:m_water * c_water * T_water = m_metal * c_metal * T_metalPlugging in the known values:50 g * 4.18 J/gC * 50C = 15 g * c_metal * 5CSolving for c_metal: 50 * 4.18 * 50  /  15 * 5  = c_metalc_metal  0.558 J/gCSo, the specific heat capacity of the metal is approximately 0.558 J/gC.