posted by sam on .
90 g of water at 25C mixed with 50g of another substance at 100C result in 45C what is specific heat of that substance
(M*C*delta T)of water + (M*C*delta T) of substance = 0
90*1.0*20 + 50*C*(-55) = 0
delta T (temperature change) of water is 20 C.
delta T of unknown substance = -55 C
Solve for the unknown specific heat C.
C = 90*20/(50*55) = 0.654 cal/(g*C)