Question

The total heat transfer can be broken into two parts: heat given off and eat absorbed (if we assume the system is closed). The heat absorbed goes into the ice, to turn it into ice water the resulting ice water, to raise it to T2 The heat given off comes from . the reservoir the water in the reservoir the thermometer all of which start out at Ti Assuming the system is closed, all of the heat given off must be absorbed. The heat equation for the system can then be expressed as: Lym+ cm (T-0u (Ti -T)+ cAmA (Ti -T2)+q (5.3) where e m is the mass of ice, as previously stated mw is the mass of water mA is the mass of the reservoir and stirrer (assuming they are both made of the same material) . c is the specific heat of water . CA is the specific heat of the (aluminum) reservoir .q is the heat released by the thermometer which is equal to: (5.4)

T1 = 46

T2 = 16

Answer 1

As stated in the problem, we have to do the order of magnitude calculation, Hence it is safe to assume the mass of ice and water as 1kg.

The various values are

specific heat of water (c)= 4186 joule/(kg K)

Specific heat of aluminium (Ca)= 904 J/(kg K)

$cm(T_2-0)= 1kg\times 4186 J/(kg\ K)\times(16K)= 66.976 \ kJ$

calculate right hand side terms

$cm_w(T_1-T_2)= 1kg\times 4186 J/(kg\ K)\times((41-16)K)= 104.650 \ kJ$

$c_Am_A(T_1-T_2)= 1kg\times 904 J/(kg\ K)\times((41-16)K)= 22.6 \ kJ$

assume V=1 (again this assumption is valid as we are concerned with only order of magnitude calculation )

$q=19.3\ J/K \times (46-16)K= 579\ J$

putting these terms in equation 5.3

$\\L_f\times 1kg+66.976 kJ=104.650kJ+22.6kJ+0.579kJ\\ L_f\times 1 kg+66.976kJ/kg=127.8kJ\\ L_f\times 1kg= (127.8-66.976)kJ=60.8kJ\\ \\ L_f=60.8kJ/kg$

The order of magnitude is

$10^5 \ J /kg$