A lead block of mass 2 kg is supplied with a heat energy of 3600 J. Calculate the rise in temperature, if the sp. heat capacity of lead is 130 J kg $^{- 1}$ C $^{- 1}$

Open in App

Solution

$Q = m \times S \times t \Rightarrow t = \frac{Q}{m \times S} = \frac{3600 J}{2 k g \times 130 J k g^{- 1}^{o} C^{- 1}}$
$\Rightarrow t = \frac{3600}{260}^{o} C = {13.85}^{o} C$

Suggest Corrections

Similar questions

1300 J

of heat energy is supplied to raise the temperature of

0.5 k g

of lead from

20^{\circ} C

40^{\circ} C

. Calculate the specific heat capacity of lead.

Q. Specific heat capacity of lead is

120 J k g^{- 1} K^{- 1}

. When

7200 J

of heat is supplied to

5 k g

of lead, the rise in temperature is

1300 J

of heat energy is supplied to raise the temperature of

0.5 k g

of lead from

20^{o} C

40^{o} C

. Calculate the specific heat capacity of lead

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from $20^{0} C$ to $40^{0} C$ . Calculate the specific heat capacity of lead.

660 J

heat energy is supplied to a solid of mass

275 g

. When its temperature rises by

1^{o} C

, calculate specific heat capacity of the solid :

Join BYJU'S Learning Program

A lead block of mass 2 kg is supplied with a heat energy of 3600 J. Calculate the rise in temperature, if the sp. heat capacity of lead is 130 J kg−1 C−1

Q=m×S×t⇒t=Qm×S=3600J2kg×130Jkg−1oC−1⇒t=3600260oC=13.85oC

A lead block of mass 2 kg is supplied with a heat energy of 3600 J. Calculate the rise in temperature, if the sp. heat capacity of lead is 130 J kg $^{- 1}$ C $^{- 1}$

$Q = m \times S \times t \Rightarrow t = \frac{Q}{m \times S} = \frac{3600 J}{2 k g \times 130 J k g^{- 1}^{o} C^{- 1}}$
$\Rightarrow t = \frac{3600}{260}^{o} C = {13.85}^{o} C$