A 505.74 g aluminum pan is removed from the stove and plunged into a sink filled with 18.07 L of water at 21^°C. The water temperature quickly rises to 24^°C. What was the initial temperature of the pan in ^°C?

 

We have formula for heat transfer

Q = m. c. ΔT

Where,

Q = amount of heat required (J)

m = mass of a substance given heat (kg)

c = specific heat of the substance (J / kg?C)

ΔT = increase / change in substance temperature (?C)

Thermodynamics explains the principle of energy conservation which states that energy cannot be created and destroyed, however, energy can be transformed from one form to another. the meaning

heat lost by aluminium (Q_a)= heat gain by water (Q_w)

m_a . c_a . (T₁ - T₃) = m_w . c_w . (T₃ - T₂)

Data given

m_a = 0. 50574 kg (mass of alumunium)

c_a = 900 J/kg ^oC. (spesific heat of Alumunium from google)

m_w = 18.07 L   (mass of water)

c_w = 4186  J/kg ^oC (spesific heat of Water from google)

T₂ = 21 ^oC  (initial temperature of the water after the pan plunged  into a sink filled)

T₃ = 24 ^oC (final temperature of the water after the pan plunged  into a sink filled)

T₁ = ......? <--------- we are looking for it ( initial temperature of the pan)

we have all data

so put it into the formula that we have

heat lost by aluminium (Q_a)= heat gain by water (Q_w)

m_a . c_a . (T₁ - T₃) = m_w . c_w . (T₃ - T₂)

0. 50574 x 900 x (T₁ - 24) = 18.07 x 4186 x (24 - 21)

455.166 x (T₁ - 24) = 226923.06

(T₁ - 24) = 226923.06 / 455.166

(T₁ - 24) = 498.55

T₁ = 498.55 + 24

T₁ = 522.55 ^oC

Therefore, the initial temperature of the pan is 522.55 ^oC