# Thermal Energy Gain and Loss

When 2 objects/substances are in thermal contact, there will be a net flow of thermal energy from the object/substance with higher temperature to the object/substance with lower temperature. If we assume there is no thermal energy loss to the surrounding

Thermal energy loss = Thermal energy gain
$m 1 c 1 Δ θ 1 = m 2 c 2 Δ θ 2$
Example 1:
A 0.5 kg block of aluminium at a temperature of 100 °C is placed in 1.0 kg of water at 20 °C. Assuming that no thermal energy is lost to the surroundings, what will be the final temperature of the aluminium and water when they come to the same temperature? [The specific heat capacity of water is 4200 Jkg-1K-1 and The specific heat capacity of aluminium is 900 Jkg-1K-1]

Answer:
Let's say the final temperature for both aluminium block and water = θ

For aluminium,
m1 = 0.5 kg
c1 = 900 Jkg-1K-1
∆θ1 = 100 °C - θ

For water,
m2 = 1.0kg
c2 = 4200 Jkg-1K-1
∆θ2 = θ - 20 °C
$m 1 c 1 Δ θ 1 = m 2 c 2 Δ θ 2 (0.5)(900)(100−θ)=(1)(4200)(θ−20) 45000−450θ=4200θ−84000 45000+84000=4200θ+450θ 4650θ=129000 θ= 27.7 o C$

Example 2:
What will be the final temperature if 50 g of water at 0 °C is added to 250 g of water at 90 °C?

Answer:
For water at 90 °C,
m1 = 250g
c1 = c2 = c
∆θ1 = 90 °C - θ

For water at 0 °C,
m2 = 50g
∆θ2 = θ - 0 °C = θ
$m 1 c Δ θ 1 = m 2 c Δ θ 2 m 1 Δ θ 1 = m 2 Δ θ 2 (250)(90−θ)=(50)(θ) 22500−250θ=50θ 22500=50θ+250θ 300θ=22500 θ= 22500 300 = 75 o C$

Example 3:
How much water at 10 °C is needed to cool 500 g of water at 90 °C down to 30 °C?

Answer:
For water at 90 °C,
m1 = 500g
c1 = c2 = c
∆θ1 = 90 - 30 = 60 °C

For water at 10 °C,
m2 = ?
∆θ2 = 30 - 10 = 20 °C
$m 1 c Δ θ 1 = m 2 c Δ θ 2 m 1 Δ θ 1 = m 2 Δ θ 2 (500)(60)= m 2 (20) 30000=20 m 2 m 2 = 30000 20 =1500g$