Question

. When 50
Calorie heat is supplied
to 30g of a
Substance, its tempesakura
is observed to increase loom
15°C to 40°c What is - (a) the
Specific heat of the substance,
6 its heat capacity ?​

Answer 1

Answer:

Heat capacity indicates the amount of energy (heat) necessary to supply to a unit of mass (grams in this case), to increase its temperature by 1 degree (celcius in this case).

If a material has an heat capacity of 3 calories/gram/º Celcius, it means you need

30 calories to increase 10ºCelcius, the temperature of 1 gram of material

Hope it helps ⭐

Answer 2

Answer:

The specific heat capacity of the substance is 0.067 cal/g °C.

Explanation:

We have given that,

• Heat supplied ( Q ) = 50 cal
• Final temperature ( T₂ ) = 40 °C
• Initial temperature ( T₁ ) = 15 °C
• Mass of substance ( m ) = 30 g

We have to find the specific heat capacity ( c ) of the object.

Now, we know that,

$\displaystyle{\pink{\sf\:Q\:=\:m\:c\:\Delta\:T}}$

$\displaystyle{\implies\sf\:Q\:=\:m\:\times\:c\:\times\:(\:T_2\:-\:T_1\:)}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{Q}{m\:\times\:(\:T_2\:-\:T_1\:)}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{Q}{m\:T_2\:-\:m\:T_1}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{50\:cal}{30\:\times\:40\:g\:^{\circ}C\:-\:30\:\times\:15\:g\:^{\circ}C}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{50\:cal}{1200\:-\:450\:g\:^{\circ}C}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{\cancel{50}\:cal}{\cancel{750}\:g\:^{\circ}C}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{\cancel{2}\:cal}{\cancel{30}\:g\:^{\circ}C}}$

$\displaystyle{\implies\sf\:c\:=\:\dfrac{1\:cal}{15\:g\:^{\circ}C}}$

$\displaystyle{\implies\sf\:c\:=\:0.0666\:cal\:/\:g\:^{\circ}C}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:c\:\approx\:0.067\:cal\:/\:g\:^{\circ}C}}}}$

∴ The specific heat capacity of the substance is 0.067 cal/g °C.