Calculating the total heat capacity of the calorimeter?

Question

50ml of water at 49.6ºC were mixed with 50ml of water at 25.1ºC in a calorimeter also 25.1ºC. the final temperature was 30.1ºC. assuming that neither the density of water nor its specific heat capacity change with temperature, calculate the total heat capacity of the calorimeter. density of water=1.00g/ml, specific heat capacity=4.18J/gK

Answer 1

we write that the amount of heat lossed by hot water is equal to the amount of heqt received by calorimeter + cold water

So Q lost by hot water is Q1 = 50 * 4.18 * (30.1-49.6)

Q received by cold wter and calorimeter
Q2 = (50+x) *4.18*(30.1-25.1) at equibrium Q1 = -Q2

-50*4.18*19.5 = -5*4.18(50+x) =20.9(50+x)

4075.5 =1045+20.9 x

3030.5=20.9x

x=145g

the calorimeter is equivalent to 145g of water

145*4.18 =606 J/°