Can you explain me how to solve this........

Question

15.1 g aluminium block is warmed to 53.2 °C and plunged into an insulated beaker containing 32.6 g of water initially at 24.4 °C. The aluminium and water are slowed to come to thermal equilibrium. Assuming that no heat is lost, what's the final temperature of the water and aluminum? 

 

How do I use these equations:

 

M_w*Cp_w*(t_f-T_w)+M_al*Cp_al-(T_f-T_al)=0

 

Water: Cpw = 4.184j/g°c

 

Aluminum: Cpal= 0.903 j/g°c

 

 

 

 

Answer 1

The warmer aluminum block will lose energy to the water until the temperature of each substance equalizes.  The thermal energy gain or loss depends on the mass (m), specific heat capacity at constant pressure (c_p), and the temperature change, ΔT.

 

 

-Q_Al = Q_w

 

 

-m_Alc_pAlΔT_Al = m_wc_pwΔT_w

 

 

(15.1 g)(0.903 J/g·°C)(53.2°C - T_f) = (32.6 g)(4.184 J/g·°C)(T_f - 24.4°C)

 

 

725.4 - 13.6T_f = 136.4T_f - 3328.1

 

 

4053.5 = 150T_f

 

 

T_f = 27.0°C