Question

Calculate the mass of ice needed to cool 150 grams of water contained in a calorimeter of mass 50 grams at 32 degrees Celsius such that the final temperature is 5 degrees Celsius. The specific heat capacity of the calorimeter is 0.4 joules per gram degree Celsius, the specific heat capacity of water is equal to 4.2 joules per gram degrees Celsius. Latent heat capacity of ice is equal to 330 joules per gram.

Answer 1

Hint: Calculate the heat loss by water + calorimeter and the heat gained by the ice. Then calculate the heat gained by the ice. Use the law of conservation of heat or equate the heat gained by the ice and the heat gained by the water and calorimeter to get the mass of ice required.

Complete step-by-step answer:
Heat loss by water + calorimeter = heat gained by the ice
We know the formula for heat transfer is given by = $Q=mS\mathrm{\Delta }T$
$H_{gained}$ by ice = $Hea{t}_{lost}$ by the calorimeter and the water.
Hence we get the heat gained by water + calorimeter : $m_w{S}_w\mathrm{\Delta }T+m_c{S}_c\mathrm{\Delta }T$
The Heat lost by ice= $m_{i}L+m_i{S}_w\delta T$
Where $m_w=50g$
$S_w=4.2{\displaystyle \frac{J}{g^{o}C}}$
$S_c=0.4{\displaystyle \frac{J}{g^{o}C}}$
$\mathrm{\Delta }T=32-5={27}^{o}C$
$m_i$ = mass of the ice
L = latent heat capacity of the ice = 330 ${\displaystyle \frac{J}{g}}$
$\delta T$ = rise in temperature = $5^{o}C$
Substituting we get the mass of the ice as $m_i={\displaystyle \frac{m_w{S}_w\mathrm{\Delta }T+m_c{S}_c\mathrm{\Delta }T}{L+S_w\delta T}}$
We get $m_i={\displaystyle \frac{150\times 4.2\times 27+50\times 0.4\times 27}{330\times 4.2\times 5}}=50g$
Thus we found the mass of the ice using the law of conservation of heat energy as 50g.

Note: One of the possible mistakes that can be made in this kind of problem is that the student directly uses the law of method of mixtures and doesn't consider the case of the melting of ice. So we need to consider the case of the possibility of melting of ice completely and calculate the amount of heat gained.