A 0.200-kg glass cup at 20°C is filled with 0.40 kg of hot water at 90°C. Neglecting any heat losses to the en?
A 0.200-kg glass cup at 20°C is filled with 0.40 kg of hot water at 90°C. Neglecting any heat losses to the environment, what is the equilibrium temperature of the water?
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Assuming no losses, the glass and water will end up at the final temperature T₂
Specific Heat from link
C_glass = 0.20
C_water = 1.00
Q_glass + Q_water = 0
M_glass C_glass ΔT_glass + M_water C_water ΔT_water = 0
200g * 0.2 * ΔT_glass + 400g * 1 * ΔT_water = 0
40 ΔT_glass + 400 ΔT_water = 0
40 (T₂ - T₁_glass) + 400 (T₂ - T₁_water) = 0
40 T₂ - 40 T₁_glass + 400 T₂ - 400 T₁_water = 0
440 T₂ = 40 T₁_glass + 40 T₁_water
T₂ = (40 T₁_glass + 400 T₁_water)/440
= (40 * 20°C + 400 * 90°C)/440
= 83.6°C
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