Question

In: Physics

Suppose I pour 13.5 mL of creamer into 159 mL of hot coffee. The creamer is...

Suppose I pour 13.5 mL of creamer into 159 mL of hot coffee. The creamer is at 43.4 deg F, and the coffee is at 188 deg F. What is the final temperature when the system reaches equilibrium, in deg F? Assume coffee and creamer have the same density and thermal properties of water.

Expert Solution

Let the final temperature of the system be 'T', then

according to heat balance,

the heat lost by coffee is equal to the heat gained by the creamer

Given, cp_coffee = cp_creamer = cp_water = c (say)

cp represents the specific heat

and, coffee = creamer = water = (say)

represents density

initial temperature of coffee = 188 F = 359.817 K

initial temperature of creamer = 43.4 F = 279.483 K

heat lost by coffee, Q_coffee = m_coffee*c_pcoffee*dT_coffee = V_coffee**c*(359.817-T) = 159**c*(359.817-T)

heat lost by creamer, Q_creamer = m_creamer*c_pcreamer*dT_creamer = V_creamer**c*(T-279.483) = 13.5**c*(T- 279.483)

According to heat balance

Q_coffee = Q_creamer

159**c*(359.817-T) = 13.5**c*(T- 279.483)

57210.903 - 159T = 13.5T - 3773.02

T = 353.53 K

OR T = 176.683 degree F

genius_generous answered 3 months ago

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