Question

Help me with the question b.​

Answer 1

Answer:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Liquid P - $Q = 3840\,J$, Liquid Q - $Q = 5500\,J$, Liquid R - $Q = 7800\,J$, Liquid S - $Q = 2856\,J$

Explanation:

a) The specific heat capacity means the amount of heat needed by a unit mass of a material to increase its temperature in one unit.

b) Let suppose that heat transfer rates between liquids and surroundings are stable. The quantity of the heat released is determined by the following expression:

$Q = m\cdot c\cdot (T_{r} - T_{f})$ (1)

Where:

$m$ - Mass of the liquid, in kilograms.

$c$ - Specific heat capacity, in joules per kilogram-degree Celsius.

$T_{r}$ - Initial temperature of the sample, in degrees Celsius.

$T_{f}$ - Freezing point, in degrees Celsius.

Liquid P ($m = 1\,kg$, $c = 160\,\frac{J}{kg\cdot ^{\circ}C}$, $T_{r} = 30\,^{\circ}C$, $T_{f} = 6\,^{\circ}C$)

$Q = (1\,kg)\cdot \left(160\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 6\,^{\circ}C)$

$Q = 3840\,J$

Liquid Q ($m = 1\,kg$, $c = 220\,\frac{J}{kg\cdot ^{\circ}C}$, $T_{r} = 30\,^{\circ}C$, $T_{f} = 5\,^{\circ}C$)

$Q = (1\,kg)\cdot \left(220\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 5\,^{\circ}C)$

$Q = 5500\,J$

Liquid R ($m = 1\,kg$, $c = 300\,\frac{J}{kg\cdot ^{\circ}C}$, $T_{r} = 30\,^{\circ}C$, $T_{f} = 4\,^{\circ}C$)

$Q = (1\,kg)\cdot \left(300\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 4\,^{\circ}C)$

$Q = 7800\,J$

Liquid S ($m = 1\,kg$, $c = 102\,\frac{J}{kg\cdot ^{\circ}C}$, $T_{r} = 30\,^{\circ}C$, $T_{f} = 2\,^{\circ}C$)

$Q = (1\,kg)\cdot \left(102\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (30\,^{\circ}C - 2\,^{\circ}C)$

$Q = 2856\,J$