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 $65 ° C$ ? (Assume that the ice is initially $65 ° C$ . The specific heat capacity of ice is $0.5 cal / g ° C$ .

Step-by-Step Solution

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Answer

Short Answer:

$m = 45.9 g$

1Given Information:

Mass of cup = 200 g

Initial Temp = 65 ^oC

2Step 1:

We have a 200 gram cup of boiling tea that we'd want to chill down to $65 ° C$ before we drink it, by putting a mass m of ice (initially at -15 ^oC ) into the tea Given that the specific heat $cal \cdot g^{- 1} \cdot K^{- 1}$ . Assume that the tea have the same heat capacity as pure water 1 $cal \cdot g^{- 1} \cdot K^{- 1}$ , the tea must decrease by $35 K$ so it must give up an amount of heat:

$Q_{neater} = m c Δ T$

$Q_{tea} = 200 \times 1 \times (35) - 7000 cal$

This heat goes into, first, heating the ice by 15 degrees to its melting point, then melting it, then heating the resulting water to 65 ^oC.

3Step 2:

For the first step, the required heat raises the temperature of ice to its melting point is:

$Q_{1} = m c Δ T$

where m is the mass of the ice, c is the specific heat of ice $0.5 cal \cdot g^{- 1} \cdot K^{- 1}$ , and $∆ T$ is the temperature difference between the initial temperature of ice and the melting point, so

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

In the second step, the amount of heat required to melt the ice is:

$Q_{2} = m \cdot L$

where L is the latent heat, and it's $80 cal / g$ for melting ice, so:

$Q_{2} = m \cdot 80 = 80 mcal$

- In the third step, the amount of heat lost by the tea to make the resulting water temperature at $65 ° C$ is,

$Q_{3} = m c Δ T$

where m is the mass of the melted ice (water), c is the specific heat of water $1 cal \cdot g^{- 1} \cdot K^{- 1}$ and $∆ T$ is the temperature difference between the initial temperature of melted ice and the final temperature of the mixture at $65 ° C$ , so: