Question

200 g mass of certain metal of 83°C is immersed in 300 g of water at 30°C the final temperature
is 33°C. Calculate the specific heat capacity of the metal Assume that the specific heat capacity
of water is `4.2 J g^-1K^-1`

Answer 1

Mass of certain metal(m₁) = 200 g
Temperature (T₁) = 83°C
Mass of water(m₂) = 300 g
Temperature of water (T₂) = 30°C
Final temperature (T) = 33°C

Specific heat capacity of water c₂ = 4.2 J/g/K
The specific heat capacity of the metal c1 is given by this formula

`m_1 c _ 1 ( T _ 1 - T) = m_2 c_2 (T - T _ 2)`

`((T - T _ 2)m_2 c_2)/(m_1(T_1 - T) ) = c_1`

`c_1 =((33 - 30) xx 300 xx 4.2)/(200 xx (83 - 33))`

`c_1 = 0.378JK ^-1`