Question

You pour 210 g hot coffee at 78.7 degree C and some cold cream at 7.50 degree C to a 115-g cup that is initially at a temperature of 22.0 degree C. The cup, coffee, and cream reach an equilibrium temperature of 63.0 degree C. The material of the cup has a specific heat of 0.2604 kcal/(kg middot degree C) and the specific heat of both the coffee and cream is 1.00 kcal/(kg middot C). If no heat is lost to the surroundings or gained from the surroundings, how much cream did you add? g An ice bag containing 0 degree C ice is much more effective in absorbing heat than one containing the same amount of 0 degree C water. The specific heat capacity of water is 1.00 kcal/(kg middot degree C), and its latent heat fusion is 79.8 kcal/kg. How much heat in kcal is required to raise the temperature of 0.740 kg of water from to 25.0 degree C?

Answer 1

The first problem have lost information.

22.- the energy added or removed from a material with a temperature change is calculated as

$\small Q = mc\Delta T$

where $\small m$ is the mass of material, $\small c$ the specific heat, and $\small \Delta T$ the temperature change during the energy change

the zero law of thermodynamics stablidhes that the energy flow is from hot objects to cold objects, and the conservation of energy say us,

$\small Q_{cold} = -Q_{hot}$

the energy lost by the hot object is gained by the cold object.

In this case the coffe lost energy and the creap and cup gain it. that is

$\small -Q_{coffe} = Q_{cream} + Q_{cup}$

$\small -m_{coffe}c_{coffe}\Delta T_{coffe} = m_{cream}c_{cream}\Delta T_{cream} + m_{cup}c_{cup}\Delta T_{cup}$

We know all that variables except the mass of the cream, then solving for that we have

$\small m_{cream} = \frac{-m_{coffe}c_{coffe}\Delta T_{coffe} - m_{cup}c_{cup}\Delta T_{cup}}{c_{cream}\Delta T_{cream}}$

$\small m_{cream} = \frac{-\left ( 0.210\textup{ kg} \right )\left ( 1.00\textup{ kcal/kg}^\circ \textup{C} \right )\left ( 63^\circ \textup{C} - 78.7^\circ \textup{C} \right ) - \left ( 0.115\textup{ kg} \right )\left ( 0.260\textup{ kcal/kg}^\circ \textup{C} \right )\left ( 63^\circ \textup{C} - 22.0 ^\circ \textup{C} \right )}{1.00 \textup{kcal/kg} ^\circ \textup{C}\left (63^\circ \textup{C} - 7.50^\circ \textup{C} \right )}$

$\small m_{cream} = \frac{3.297\textup{ kcal} - 1.2259\textup{ kcal}}{55.5 \textup{kcal/kg}}$

$\small m_{cream} = 0.0373\textup{ kg}$

$m_{cream} = 37.3\textup{ g}$

23.- The heat required to raise the temperature of water from 0°C to 25°C is

$Q = mc\Delta T$

$Q = \left (0.740\textup{ kg} \right )\left ( 1.00\textup{ kcal/kg} ^\circ\textup{C} \right )\left ( 25^\circ \textup{C} - 0^\circ \textup{C}\right )$

$Q = 18.5\textup{ kcal}$