Question

10. 360 J of heat is supplied to rise the temperature of a certain mass of iron from 20 °C to 25 °C. If the
specific het of iron is 470 JKg-1 °C then t he mass of iron is
a) 15 kg
b) 1.5 kg
c) 0.15 kg
d) 0.015 kg​

Answer 1

Given :-

• Heat energy = 360J

• Initial temperature of iron, $\sf t_1 = 20 ^{\circ}C$

• Final temperature of iron, $\sf t_2 = 25^{\circ}C$

• Specific heat capacity of iron = $\sf 470 \ JKg^{-1} \ ^{\circ}C$

To Find :-

• Mass of iron.

Solution :-

Let mass of iron be m.

$\underline {\boxed{\bf Specific \ heat \ capacity = \dfrac{Heat \ energy }{Mass \times Rise \ in \ temperature}}}$

Here :-

• Heat energy = 360J
• Rise in temperature = $\sf t_2-t_1$

                                          = $\sf 25 - 20$

                                          = $\sf 5 ^{\circ}C$

• Specific heat capacity = $\sf 470 \ JKg^{-1} \ ^{\circ}C$

• Mass = m

$\longrightarrow \sf 470 = \dfrac{360}{5 \times m } \\\\\longrightarrow 5m = \dfrac{360}{470} \\\\\longrightarrow 5m = 0.766 \\\\\longrightarrow m = \dfrac{0.766}{5} \\\\\longrightarrow \bf m = 0.15$

Mass of iron = 0.15 kg

Answer 2

Answer:

$Solution :- \\ </p><p></p><p>Let mass of iron be m. \\ </p><p></p><p> \underline{ \boxed{\mathbb{\pink{\underline {\boxed{\bf Specific \ heat \ capacity = \dfrac{Heat \ energy }{Mass \times Rise \ in \ temperature}}}}}}}\\ \\ { \boxed{\mathbb{\green{pecific heat capacity=Mass×Rise in temperatureHeat energy}}}} \\ </p><p></p><p> \underline{ \boxed{\mathbb{\red{Here :- }}}}\\ \\ </p><p></p><p>Heat energy = 360J \\ Rise in temperature = \sf t_2-t_1t2−t1 \\ </p><p></p><p>                                          = \sf 25 - 2025−20 \\ </p><p></p><p>                                          = \sf 5 ^{\circ}C5∘C \\ </p><p></p><p>Specific heat capacity = \sf 470 \ JKg^{-1} \ ^{\circ}C470 JKg−1 ∘CMass = m \\ </p><p></p><p>\begin{gathered}\longrightarrow \sf 470 = \dfrac{360}{5 \times m } \\\\\longrightarrow 5m = \dfrac{360}{470} \\\\\longrightarrow 5m = 0.766 \\\\\longrightarrow m = \dfrac{0.766}{5} \\\\\longrightarrow \bf m = 0.15\end{gathered} \\ ⟶470=5×m360 \\ ⟶5m=470360 \\ ⟶5m=0.766 \\ ⟶m=50.766 \\ ⟶m=0.15 \\ \\ { \boxed{\mathbb{\blue{m = 0.15}}}}</p><p></p><p>$