A 30 gram metal at 100 degrees Celsius was dropped into 132 grams of water at 25 degrees Celsius and the final temperature for the metal in water system was 29 degrees Celsius. Calculate the specific heat capacity of the metal if the specific heat capacity of water is 4.2J/G/C. (assume a closed system)

Your school subject is probably physics. Your subject is definitely not the name of your university.

To calculate the specific heat capacity of the metal, we can use the principle of conservation of energy. The energy lost by the metal is gained by the water.

First, let's find the energy gained by the water:
q_water = m_water * c_water * ΔT_water

Where:
m_water = mass of water = 132 grams
c_water = specific heat capacity of water = 4.2 J/g/°C
ΔT_water = change in temperature of water = final temperature - initial temperature
= 29°C - 25°C = 4°C

Substituting the values:
q_water = 132 g * 4.2 J/g/°C * 4°C
q_water = 2,646 J

According to the principle of conservation of energy, the energy gained by the water is equal to the energy lost by the metal. So, we can find the specific heat capacity of the metal using the equation:

q_metal = m_metal * c_metal * ΔT_metal

Where:
m_metal = mass of metal = 30 grams
c_metal = specific heat capacity of the metal (what we want to find)
ΔT_metal = change in temperature of the metal = final temperature - initial temperature
= 29°C - 100°C = -71°C

Substituting the values:
2,646 J = 30 g * c_metal * -71°C
c_metal = -2,646 J / (30 g * -71°C)
c_metal = 1.56 J/g/°C

Therefore, the specific heat capacity of the metal is 1.56 J/g/°C.