Question

A chemistry student wants to determine the specific heat capacity of a metal sample. The student adds 50 g of the metal sample to 200 g of water at 25°C in a calorimeter. The temperature of the water rises from 25°C to 38°C. Assume the specific heat capacity of water is 4.184 J/g°C. What is the specific heat capacity of the metal sample?

Answer 1

To determine the specific heat capacity of the metal sample, we can use the formula for heat transfer:q = mcTwhere q is the heat transferred, m is the mass, c is the specific heat capacity, and T is the change in temperature.First, we need to find the heat gained by the water:q_water = m_water * c_water * T_waterq_water =  200 g  *  4.184 J/gC  *  38C - 25C q_water = 200 * 4.184 * 13q_water = 10879.2 JSince the metal sample loses heat to the water, the heat gained by the water is equal to the heat lost by the metal:q_metal = -q_waterq_metal = -10879.2 JNow we can use the heat transfer formula to find the specific heat capacity of the metal:q_metal = m_metal * c_metal * T_metal-10879.2 J =  50 g  * c_metal *  25C - 38C -10879.2 J = 50 * c_metal *  -13C Now, we can solve for c_metal:c_metal = -10879.2 J /  50 g * -13C c_metal = 10879.2 / 650c_metal  16.74 J/gCThe specific heat capacity of the metal sample is approximately 16.74 J/gC.