Question

If you pour whisky over ice, the ice will cool the drink, but it will also dilute it. A solution is to use whisky stones. Suppose Ernest pours 55.0 g of whisky at 22 ∘C room temperature, and then adds three whisky stones to cool it. Each stone is a 32.0 g soapstone cube that is stored in the freezer at -11 ∘C. The specific heat of soapstone is 980 J/kg⋅K; the specific heat of whisky is 3400 J/kg⋅K.What is the final temperature of the whisky?

Answer 1

Answer:

10.1 °C

Explanation:

After the mixture, heat is gained by the soapstone cubes (they were at a lower temperature) while heat is lost by the whisky (it was at a higher temperature).

Let the temperature of the mixture be T.

Heat lost by whisky, $H_W = m_Wc_W\Delta\theta_W$

$m_W$ is the mass of the whisky (in kg), $c_W$ is the specific heat of whisky (J/kg·K) and $\Delta\theta_W$ is the change in temperature of the whisky (in °C or K).

$H_W = (0.055\text{ kg})(3040\text{ J/kg}\cdot\text{K})(22-T) = 167.2(22-T) \text{ J}$

Heat gained by soapstone cubes, $H_C = m_Cc_C\Delta\theta_C$

There are three cubes of soapstone, so their mass is multiplied by 3.

$H_W = (3\times0.032\text{ kg})(980\text{ J/kg}\cdot\text{K})(T-(-11)) = 94.08(T+11) \text{ J}$

From the principle of mixtures,

Heat lost by whisky = Heat gained by soapstone cubes (assuming no heat loss to the surroundings)

$167.2(22-T) = 94.08(T+11)$

$3678.4 - 167.2T = 94.08T + 1034.88$

$2643.52 = 261.28T$

$T= 10.1$

The final temperature of the whisky is 10.1 °C.

Answer 2

Answer: 11.0

Explanation: the heat capacity of the whiskey should be 3400 and not 3040