A cube of tin with a mass of 1.90 kg, initially at a temperature of 150.0°C, is in a well-insulated container. Water at a temperature of 29.0°C is added to the container, and the entire interior of the container is allowed to come to thermal equilibrium, where it reaches a final temperature of 64.0°C. What mass of water (in kg) was added? Assume any water turned to steam subsequently recondenses. Answer is not 0.23kg
given
m_tin = 1.90 kg
we know, specific heat of tin, C_tin = 210 J/(kg K)
specific heat of water, C_water = 4180 J/(kg K)
let m is the mass of water
Apply, heat gained by water = heat lost by tin
m_water*C_water*(64 - 29) = m_tin*C_tin*(150 - 29)
m_water = m_tin*C_tin*(150 - 29)/(C_water*(64 - 29) )
= 1.90*210*121/(4186*35)
= 0.330 kg <<<<<<<---------------------Answer
A cube of tin with a mass of 1.90 kg, initially at a temperature of 150.0°C,...