Question

A hot solid of mass 60 g at 100${}^{\circ }C$ is placed in 150 g of water at 20${}^{\circ }C$. The final temperature recorded is 25${}^{\circ }C$. Calculate the specific heat of solid. (Specific heat of water = 4200J/kg/${}^{\circ }C$)

• A. 50J/kg/⁰C
• B. 100J/kg/⁰C
• C. 300J/kg/⁰C
• D. 700J/kg/⁰C

Answer 1

The correct option is D

700J/kg/⁰C

Water Solid

$m_1=150g$ $m_2=60g$

$c_1=4200J/kg{/}^{0}C$ $c_2=?$

$t_1={20}^{0}C$ $t_2={100}^{0}C$

T = final temperature of mixture = ${25}^{0}C$

By the principle of calorimetry

Heat lost of solid = Heat gain of solid

$m_2{c}_2\Delta t_2=m_1{c}_1\Delta t_1$

$60gm\times c_2(100–25)=150g\times 4200J/k{g}^{0}C(25–20)$

$60c_2\times 75=150\times 4200\times 5$

$c_2=\frac{150\times 4200\times 5}{60\times 75}=700J/kg{/}^{0}C$