Q. 1.47

Question

Your $200 g$ cup of tea is boiling-hot. About how much ice should you add to bring it down to a comfortable sipping temperature of $65^{\circ} C$ ? (Assume that the ice is initially at $- 15^{\circ} C$ . The specific heat capacity of ice is $(0.5 cal / g \cdot^{\circ} C .)$ .)

Step-by-Step Solution

Verified

Answer

The sum of these three heat is equal to the heat lost by the tea $m = 45.9 g$

1Step : 1 Thermal capacity

We have a $200$ gram cup of boiling tea and want to chill it down to $65^{o} C$ before drinking it, so we put a mass $m$ of ice (at $- 15^{O} C$ ) into it. Given that ice has a thermal capacity of Assuming that the tea has the same heat capacity as pure water $1 cal \cdot g^{- 1} \cdot K^{- 1}$ , the tea's heat capacity must fall by $35^{\circ}$ , implying that it must give up some heat:

$Q_{water} = m c Δ T$

$Q_{t e a} = 200 \times 1 \times (35) = 7000 cal$

2Step : 2 Stages of melting ice

The required heat to raise the temperature of ice to its melting point in the first phase is:

$Q_{1} = m c Δ T$

where $m$ is the ice's mass, $c$ is the ice's specific heat $0.5 cal \cdot g^{- 1} \cdot K^{- 1}$ is the temperature difference between the ice's original temperature and the melting point, so:

$Q_{1} = m 0.5 \times (0 - (- 15)) = 7.5 m cal$

The amount of heat necessary to melt the ice in the second stage is:

$Q_{2} = m \times L$

where $L$ is the heat $80 cal / g$ for melting ice

$Q_{2} = m \times 80 = 80 m cal$

The amount of heat lost by the tea in the third phase to achieve a water temperature of $65^{0} C$ is,

$Q_{3} = m c Δ T$

where $m$ is the mass of melted ice (water), and $c$ is water's specific heat. $1 cal \cdot g^{- 1} \cdot K^{- 1} and Δ T$ is the temperature difference between the beginning temperature of melting ice and the ultimate temperature of the combination at $65^{0} C$ .