Question

Suppose that 100g of aluminum at 23 degrees C is added to 50g of water at 95 degrees C. Determine the equilibrium temperature of the mixture. Ignore heat loss to the container or the surroundings.

Answer 1

heat capacities of both the substances be given in such situations

Answer 2

The equilibrium temperature is 73.34 °C

Step-by-step explanation:

Step 1: Data given

Mass of aluminium = 100.0 grams

Initial temperature of aluminium = 23.0 °C

Specific heat of aluminium = 0.900 J/g°C

Mass of water = 50.0 grams

Initial temperature of water = 95.0 °C

Specific heat of water = 4.184 J/g°C

Step 2: Calculate equilibrium temperature

Heat gained = heat lost

Q = m*c*ΔT

Qaluminium = - Qwater

m(aluminium) * c(aluminium) * ΔT(aluminium) = -m(water) * c(water) * ΔT(water)

⇒mass aluminium = 100.0 grams

⇒ c(aluminium) = specific heat aluminium = 0.900 J/g°C

⇒ΔT = Change in temperature of aluminium = T2 - T1 = T2 - 23.0 °

⇒mass water = 50.0 grams

⇒ cwater) = specific heat ater = 4.184 J/g°C

⇒ΔT = Change in temperature of water = T2 - T1 = T2 - 95.0 °

100.0g * 0.900 J/g°C * (T2- 23.0°C) = - 50.0g * 4.184J/g°C *(T2- 95.0°C)

90*(T2-23.0°C) = -209.2(T2- 95.0°C)

90T2 -2070 = -209.2T2 + 19874

299.2T2 = 21944

T2 = 73.34 °C

The equilibrium temperature is 73.34 °C