Question

A porcelain cup of mass 303 g and specific heat 0.260 cal/g-°C contains 161 cm³ of coffee, which has a specific heat of 1.00 cal/g-°C. If the coffee and cup are initially at 71.0 °C, how much ice at 0.00 °C must be added to lower the temperature to 49.0 °C?

Answer 1

Answer:

$m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr$

So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C

Explanation:

For this case we need to use the fact that the sum for all the heats involved in the system are 0, since we assume an equilibrium state.

Data given

$m_p = 303 gr$ mass of the porcelain cup

$cp_p = 0.260 cal/g C$ the specific heat for the porcelain cup

$T_{ip} = T_{ic}= 71 C$ initial temperature for the coffee and the porcelain cup.

$V_{c}= 161 cm^3$ Volume of the coffee.

We can convert this to m^3 and we got 0.000161m^3 and assuming the density fot the coffee equal to the water 1 Kg/m^3 the mass would be:

$m_c = 1 kg/m^3 *0.000161 m^3 = 0.000161 kg=0.161 Kg$

$Cp_{c} = 1 cal/g C$ Specific heat for the coffee

$m_i =?$ mass of ice required

$T_e= 49C$ equilibrium temperature

$L_f = 80 cal/g$ represent the latent heat of fusin since the ice change the state to liquid.

Solution to the problem

Using this formula:

$\sum_{i=1}^n Q_i = 0$

We have this:

$m_p cp_p (T_e -T_{ip}) + m_{c} cp_c (T_e -T_{ic}) +m_i L_f + m_i cp_w (T_e -0) =0$

Now we can replace and we have this:

$303 gr *(0.260 cal/g C) (49-71)C + 0.161 gr*(1 cal/g C)(49-71)C +m_i [80 cal/gr+(1cal/g C)(49-0)C]=0$

And now we can solve for $m_i$ and we have:

$-1733.16cal -3.542cal +m_i [129 cal/g]=0$

$m_i =\frac{1736.702 cal}{129 cal/gr}=13.46 gr$

So we need to add 13.46 gr of ice in order to reach the final equilibrium temperature of 49 C