Question

120 grams of boiling water (temperature 100° C, heat capacity 4.2 J/gram/K) are poured into an aluminum pan whose mass is 1100 grams and initial temperature 22° C (the heat capacity of aluminum is 0.9 J/gram/K).
(a) After a short time, what is the temperature of the water?
Tfinal = C
(b) What simplifying assumptions did you have to make?
-Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.
-The heat capacities for both water and aluminum hardly change with temperature in this temperature range.
-The thermal energy of the aluminum doesn't change.
-The thermal energy of the water doesn't change.
(c) Next you place the pan on a hot electric stove. While the stove is heating the pan, you use a beater to stir the water, doing 26432 J of work, and the temperature of the water and pan increases to 77.8° C. How much energy transfer due to a temperature difference was there from the stove into the system consisting of the water plus the pan?
Q = J

Answer 1

Answer:

Part a)

$T = 48.35 ^oC$

Part b)

-Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.

-The heat capacities for both water and aluminum hardly change with temperature in this temperature range.

Part c)

$Q_{net} = 70430.3 J$

Explanation:

Part a)

Heat given by water = heat absorbed by container

so we will have

$m_ws_w\Delta T_1 = m_c s_c \Delta T_2$

now we have

$120 (4.2)(100 - T) = 1100(0.9)(T - 22)$

$100 - T = 1.96(T - 22)$

$143.12 = 2.96 T$

$T = 48.35 ^oC$

Part b)

-Energy transfer between the system (water plus pan) and the surroundings was negligible during this time.

-The heat capacities for both water and aluminum hardly change with temperature in this temperature range.

Part c)

Total energy transfer due to temperature difference is given as

$Q = (m_ws_w + m_cs_c)\Delta T$

$Q = (120\times 4.2 + 1100\times 0.9)(77.8 - 48.35)$

$Q = 43998.3 J$

Also we did work of 26432 J on it so total energy transferred to it

$Q_{net} = 43998.3 + 26432$

$Q_{net} = 70430.3 J$

Answer 2

(a) The final temperature of the water in the mixture after a short time is 48.31 ⁰C.

(b) The simplifying assumption to take is energy transfer between the system (water plus pan) and the surroundings was negligible during this time.

(c) The total energy transferred due to the temperature difference from the stove into the system is 70,490.1 J.

The given parameters;

• mass of the water, $m_w$ = 120 g
• temperature of the water, $t_w$ = 100° C
• heat capacity of the water, $c_w$ = $4.2 \ J/gk$
• mass of the aluminum pan, $m_a$ = 1100 g
• initial temperature of the aluminum, $t_a$ = 22° C
• heat capacity of the aluminum, $c_a$ = $0.9 \ J/gk$

The final temperature of the water is obtained by applying the principle of mixtures;

$Q_{H_2O} = Q_{Al}$

$m_wc_w (100-t_f) = m_ac_a (t_f - 22)\\\\120\times 4.2\times (100-t_f) = 1100\times 0.9(t_f-22)\\\\50,400 - 504t_f = 990t_f -21,780\\\\72,180 = 1494t_f\\\\t_f = \frac{72,180}{1494} \\\\t_f = 48.31 \ ^0C$

(b)

The simplifying assumption to take is energy transfer between the system (water plus pan) and the surroundings was negligible during this time.

(c)

The total heat gained by the mixture as a result of the temperature difference is calculated as;

$Q_t = Q_w + Q_a\\\\Q_ t = m_wc_w\Delta t + m_ac_a\Delta t\\\\Q_t = \Delta t(m_wc_w \ + \ m_ac_a)\\\\Q_t = (77.8 - 48.31) (120\times 4.2 \ \ + \ \ 1100\times 0.9)\\\\Q_t = 44,058.1 \ J$

The total energy transferred is calculated as;

$E = Q_t + W\\\\E = 44,058.1 + 26,432\\\\E = 70,490.1 \ J$

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