Answer to Question #164034 in Physics for Dozzy B

Question

Dry steam is passed into a well lagged copper can of mass 250g containing 400g of water and 50g of ice at 0degreeC. The mixture is well stirred and the steam supplied is cut off when the temp of the can and its content reaches 20degreeC.Neglecting heat losses, find the mass of steam condensed. [Specific heat capacity of water=4.2jg-1k-1, specific heat capacity of copper=0.4jg-1k-1, specific latent heat of steam=2260j/g, specific latent heat of ice=336j/g]?

Answer 1

$$m_sR=m_iL+(m_w+m_i)c_w\Delta T+m_cc_c\Delta T\\ 2260m_s=50\cdot336+(400+50)(4.2)(20)\\+(250)(0.4)(20)\\m_s=25\ g$$