Solution:

Given the information in the question:

Mass of the metal block, m = 0.20 kg = 200 g
The initial temperature of the metal block, T₁​ = 150⁰C
The final temperature of the metal block, T_2​ = 40⁰C
Copper calorimeter has water equivalent of mass, m₁= 0.025 kg = 25 g
Volume of water, V = 150 cm³
Mass (M) of water at temperature T = 27⁰C is $150\times 1=150g$

Specific heat of water, C_w$=4.186J/g/K$
Let Specific heat of the metal be c

Evaluating the decrease in the temperature of the metal block ΔT₁=T_1​–T_2​ =150−40=110^oC
Thermal expansion (increase in temperature of the water and calorimeter system), ΔT₂=40−27=13⁰C

Heat lost by the metal = Heat gained by the water + heat gained by the  calorimeter

mC$\Delta$T₁ = (M+m₁)C_wΔT₂

C = [(M+m₁)C_wΔT₂]/m$\Delta$T₁

= [(150 +25) x 4.186$x\; 13]/(200\; x\; 100)$

=(175 x 4.186$x\; 13$$)/(200\; x\; 100)$

= 9523.15/22000

= 0.43 Jg⁻¹k⁻¹

Thus, answer is 0.43 Jg⁻¹k⁻¹

Unless some heat is lost to the surrounding environment, the specific heat of metal will be less than its true value.