Question

2) In a calorimetry experiment, 70.0 g of a substance was heated to

Answer 1

The temperature at thermal  equilibrium : 16.1 °C

Further explanation

Q absorbed = Q released

Heat can be formulated :

Q = m.c.Δt

c= specific heat, J/g°C

Δt = temperature different, C°

m= mass, g

Q water=Q substance

$\tt 3500\times 4.18\times (t_2-15)=70\times 0.89\times (270-t_2)\\\\14630(t_2-15)=62.3(270-t_2)\\\\14630t_2-219450=16821-62.3t_2\\\\14692.3t_2=236271\\\\t_2=16.1^oC$