(e) The temperature of the coffee decreases as the device is used. The initial temperature of the coffee was 76^circ C The internal energy of the coffee decreased by 15 kJ. density of coffee=1.1times 10^3kg/m^3 volume of coffee=1.9times 10^-4m^3 specific heat capacity of coffee=4200J/kg^circ C Calculate the final temperature of the coffee. Use the Physics Equations Sheet.

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(e) The temperature of the coffee decreases as the device is used. The initial temperature of the coffee was 76^circ C The internal energy of the coffee decreased by 15 kJ. density of coffee=1.1times 10^3kg/m^3 volume of coffee=1.9times 10^-4m^3 specific heat capacity of coffee=4200J/kg^circ C Calculate the final temperature of the coffee. Use the Physics Equations Sheet.

Answer 1

Firstly, from step 1, using the formula \( mass = density \times volume \), we obtain the mass of coffee as: \( mass = 1.1 \times 10^{3} \mathrm{~kg/m^{3}} \times 1.9 \times 10^{-4} \mathrm{~m^{3}} = 0.209 \) kg. Next, the decrease in internal energy should be converted into Joules because the specific heat capacity is given in Joules. So, \( \Delta U = 15 \times 10^{3} \) J. Now, using step 2 formula for the final temperature: \( T_f = T_i - \Delta U/(mc) \), plugging in the numbers gives: \( T_f =76 - (15 \times 10^{3} \mathrm{~J}) / (0.209 \\xrightarrow[]{}{~kg} \times 4200 \mathrm{~J/ \xrightarrow[]{}{~kg} }^{\circ} \mathrm{C}) = 32.88^{\circ}C.\) So, the final temperature of the coffee after use is \( 32.88^{\circ}C \).

Answer 2

## Step 1: We need to first calculate the mass of the coffee. Since density is equal to mass per unit volume, we can rearrange the formula to find the mass of the coffee which is the product of density and volume. ### The formula for this step is \(mass = density \times volume\) ## Step 2: Knowing that change in internal energy equals mass times heat capacity times change in temperature (\(\Delta U = mc\Delta T \), where \( \Delta U \)= change in energy, \(m\) = mass, \(c\) = specific heat capacity, and \(\Delta T = (T_f - T_i) \) where \(T_f\) and \(T_i\) are the final and the initial temperature respectively. We've already calculated the mass of the coffee using the density and volume. The initial temperature, plug in the internal energy (notice the unit conversion to get the energy in Joules) and specific heat are given in the question. Solving for \(T_f\) in equation \(1\) yields: ### Rearranging the equation for final temperature yields \(T_f = T_i - \Delta U/ (mc) \) This is the primary formula we need to solve. ## Step 3: Plug in numbers:

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