A copper ball of mass 100g is at a temperature T . It is dropped in a copper calorimeter of mass 100g, filled with 170g of water at room temperature. Subsequently, the temperature of the system is found to be 75 ° C . T

Correct answer: C. 885 ° C cases m _ cb = 100g, & & ( T ^ o C) m _ c = 100g, & & ( 30 ^ o C) m _ w = 170g, & & ( 30 ^ o C) cases Net heat loss = Net heat gain ⇒ 100 × 0.1 × (T - 75) = 100 × 0.1 × (75 - 30) + 170 × 1 × (75 - 30) ⇒ T - 75 = 45 + 17 × 45 ⇒ T = 885 ° C .

JEE Main 20172 April

Exam: JEE Main
Year: 2017
Date: 2 April
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2017 · 2 April · Physics · Thermal Properties of Matter

A copper ball of mass 100g is at a temperature T . It is dropped in...

A copper ball of mass

100\text{g}

is at a temperature

T

. It is dropped in a copper calorimeter of mass

100\text{g,}

filled with

170\text{g}

of water at room temperature. Subsequently, the temperature of the system is found to be

75 ° C

T

is given by:
(Given: room temperature

= 30 ° C

, specific heat of copper

= 0.1cal{g}^{- 1} ° {C}^{- 1}

)

Options

A $825 ° C$
B $800 ° C$
C $885 ° C$
D $1250 ° C$

Correct answer

C. $885 ° C$

Step-by-step solution

\begin{cases}{m}_{cb} = 100g, & & \left({T}^{o}C\right) \\ {m}_{c} = 100g, & & \left({30}^{o}C\right) \\ {m}_{w} = 170g, & & \left({30}^{o}C\right)\end{cases}

Net heat loss = Net heat gain

⇒ 100 \times 0.1 \times \left(T - 75\right) = 100 \times 0.1 \times \left(75 - 30\right) + 170 \times 1 \times \left(75 - 30\right)

⇒ T - 75 = 45 + 17 \times 45

⇒ T = 885 ° C

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