A 50.0g copper calorimeter contains 250g of water at 20 deg celsius. How much steam must be condensed into the water if the final temperature of the system is to reach 50 deg celsius?
Can someone explain how to find the Ls? I was told that just plugging in the latent heat of steam (2.26*10^6) is incorrect.
The Attempt at a Solution
Qsteam = Qcalorimeter + Qwater
ms * Ls = (mc * cc * dT) + (mw * cw * dT)
ms * Ls = (0.05kg * 0.0924cal/g * (50deg-20deg)) + (.25kg * 1cal/g *(50deg-20deg))
ms * Ls = 7638.6cal