Calculate your calorimeter constant (Ccalorimeter) for Trial 1 and Trial 2.

Question

Trial 1:

Mass of cold water: 50.0

Temp of cold water: 21.6

Mass of hot water: 49.5

Temp of hot water : 50.7

Temp of mixed water: 38.2

Trial 2:

Mass of cold water: 49.9

Temp of cold water: 22.8

Mass of hot water: 49.3

Temp of hot water : 55.8

Temp of mixed water: 38.7

Answer 1

Mass of cold water: 50.0

Temp of cold water: 21.6

Mass of hot water: 49.5

Temp of hot water : 50.7

Temp of mixed water: 38.2 (Final temp for both samples)

specific heat of liquid water = 4.184 J/gC

q=C_calΔT+m_waterc_waterΔT

-C_calΔT+m_waterc_waterΔT=+C_calΔT+m_waterc_waterΔT

negative value for initially hot water, because it will cool down and is therefore exothermic (negative q)

substitute the values, where C_cal=x

x(38.2-50.7)+49.5(4.184)(38.2-50.7)=x(38.2-21.6)+50.0(4.184)(38.2-21.6)

solve for x

should be:

x≈-1478.43

which is probably Ccal=1478.43 J/^oC,

so, 1.47843 kJ/^oC

**edit: had a typo initially in wolfram alpha, fixed now