Calculate the heat lost by the hot water in the calorimeter

Question

Calculate the heat lost by the hot water in the calorimeter (qhot water).

for water use 1.00 g/mL as its density and 4.18 J/gK as its specific heat capacity.

Mass of cold water: 50.1

Temperature of cold water:22.4

Mass of hot water:50.0

Temperature of hot water:61.3

Temperature of mixed water:41.3

Answer 1

heat lost by hot water = q = mC∆T

m = mass of hot water = 55 g

C = specific heat water = 4.18 J/gº

∆T = change in temperature = 41.3º - 61.3º = -20º

q = (55 g)(4.18 J/gº)(41.3 - 61.3) = -4598 J = heat lost by hot water