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To answer this question (at least the only way I know of) you first need to know the specific heat capacities (symbolized by C) of both milk and coffee. The formula for heat transfer equations is:

q1 = -q2

where (we'll call q1 milk's heat) q1 = C (specific heat of milk) x ΔT x mass (g)

and q2 = C (specific heat coffee) x ΔT x mass (g)

and where ΔT = (final temperature - initial temperature)

So, if you substitute in 250g for the mass of coffee, 90 degrees for its initial temperature, do the same for the milk side of the equation, and put their specific heat capacities in the two equations, you can solve for the final temperature by substitution.

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14y ago
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10y ago

Assuming the specific heat of both is 1 because both are a very high percentage of water, then

x (90 - 10)(1) = 250 (95 - 90)(1)

80 x = 1250

x = 1250/80 = 15.625 grams ( ml ) of milk

but this ignores the heat absorbed by the cup itself.

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Q: A thermostat bottle contains of 250g of coffee at 90C to this is added 20g of milk at 5C After equilibrium is established what is the temperature of the liquid?
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