Question

suppose you mix 100.0 g of water at 22.6 oc with 75.0 g of water at 75.4 oc. what will be the final temperature of the mixed water, in oc?

Answer 1

The temperature of mixed water is 45.230C under the conditions stated.

How can the change in temperature when mixing water be calculated?

T(final) = (m1 T1 + m2 T2) / (m1 + m2), at which m1 and m2 are indeed the weight training of the water in the initial and 2nd canisters, T1 is the water temperature in the first container, and T2 is the water temperature in the second container, can be used to determine the final the water's temperature mixture.

Briefing:

The given parameters;

mass of the cold water, m = 100 g

initial temperature of the water, t₁ = 22.6 ⁰C

initial temperature of the hot water, t₂ = 75.4⁰ C

75 g is the hot water's mass.

specific heat capacity of water is 4.184 J/g⁰C

The mixture's final temperature is estimated as follows;

Based on the principle of conservation of energy;

Heat received by the ice water equals heat lost by the hot water.

mcΔθ₂ = mcΔθ₁

75 x 4.184 x (75.4 - T) = 100 x 4.184 x (T - 22.6)

75 x (75.4 - T) = 100 x (T - 22.6)

(75.4 - T) = 1.333(T - 22.6)

75.4 - T = 1.333T - 30.1258

75.4 + 30.1258 = 1.333T + T

105.5258 = 2.333T

T = 45.2318⁰C

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