Question

Assume all temperatures to be exact.
A 0.25-kg glass cup at 22 ∘C is filled with 0.41 kg of hot water at 95 ∘C.

Part A

Neglecting any heat losses to the environment, what is the equilibrium temperature of the water?

Express your answer using two significant figures.

T

=

  ∘C  

Answer 1

Mass of glass, M_glass = 0.25 kg = 250 g

Mass of water, M_water = 0.41 kg = 410 g

Suppose the equilibrium temperature of glass and water is T deg C.

Now, from the text book, write down the specific heats of glass and water -

C_glass = 0.20 Cal/g*K

C_water = 1.00 Cal / g*K

Under balanced condition, total quantity of heat of both glass and water is zero.

So,

Q_glass + Q_water = 0

M_glass C_glass ΔT_glass + M_water C_water ΔT_water = 0

250g * 0.2 * ΔT_glass + 410g * 1 * ΔT_water = 0

50 ΔT_glass + 410 ΔT_water = 0

50 (T - T₁_glass) + 410 (T - T₁_water) = 0

50 T - 50 T₁_glass + 410 T - 410 T₁_water = 0

460 T = 50 T₁_glass + 410 T₁_water

T = (50 T₁_glass + 410 T₁_water)/460

= (50 * 22°C + 410 * 95°C)/460

= (1100 + 38950) / 460

= 87.0 °C (Answer)