Question

You are given \(250 \mathrm{g}\) of coffee (same specific heat as water) at \(80.0^{\circ} \mathrm{C}\) (too hot to drink). In order to cool this to $\left.60.0^{\circ} \mathrm{C}, \text { how much ice (at } 0.0^{\circ} \mathrm{C}\right)$ must be added? Ignore heat content of the cup and heat exchanges with the surroundings.

Short answer

Answer: Approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).

Step-by-step solution

Write the energy conservation equation

The energy conservation equation can be written as: heat gained by the ice + heat gained by the melted ice = heat lost by the coffee

Write the formula for heat gained/lost

Heat gained/lost can be written as \(Q = mcΔT\) for a given substance, where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(ΔT\) is the change in temperature. In this case, we have three substances: ice, melted ice (water), and coffee.

Introduce the latent heat of fusion for ice

During ice melting, we must account for the amount of heat absorbed by ice to turn to water. The heat obtained from the coffee will first be used to melt the ice; the melted ice will then gain heat from the remaining heat available. The latent heat of fusion can be written as \(Q_{fusion} = mL\), where \(Q_{fusion}\) is the heat absorbed during the phase change, \(m\) is the mass of ice, and \(L\) is the latent heat of fusion.

Write the energy conservation equation with known values and variables

Considering all the steps above, we can write the equation for this problem, taking \(m_{ice}\) as the mass of ice added: \((m_{ice}L) + (m_{ice}c_wΔT_{melted\_ice}) = (m_{coffee}c_wΔT_{coffee})\) where \(c_w = 4.18 \mathrm{J/g\:C}\) is the specific heat capacity of water, \(L = 334 \mathrm{J/g}\) is the latent heat of fusion for water, \(m_{coffee} = 250 \mathrm{g}\), \(ΔT_{melted\_ice} = (60 - 0)\), and \(ΔT_{coffee} = (80 - 60)\).

Solve the equation for mass of ice

Simplify and plug in known values: \(m_{ice}(334) + m_{ice}(4.18)(60) = (250)(4.18)(20)\) \(m_{ice}(334 + 250.8) = (250)(83.6)\) \(m_{ice} = \frac{(250)(83.6)}{584.8}\) \(m_{ice} \approx 35.7 \mathrm{g}\) So, approximately \(35.7 \mathrm{g}\) of ice is needed to cool the coffee to \(60.0^{\circ} \mathrm{C}\).