Question

A 100.0 g copper block at 80.0◦C is placed in 2.50 kg of water at 25◦C. What would be the system temperature in equilibrium?

Answer 1

mass of water m_w= 2.5 kg =2.5 *1000 g=2500 g [1 kg =1000 g]

mass of copper m_cu= 100g

temperature of copper T_cu =80⁰C

temperature of water T_water =25⁰C

let the equilibrium temperature be T

specific heat capacity of copper C_cu=0.380 J^oC^-1g

specific capacity of water C_water=4.183 J^oC^-1g

heat gained by cold body $q^{water}=m_{water}C_{water}\Delta T=m_{water}C_{water}(T-T_{water})$

q_water =2500g *4.183 JC^-1g*(T-25)

heat lost by hot body $q^{copper}=m_{copper}C_{copper}\Delta T=m_{copper}C_{copper}(T_{copper}-T)$

q_copper= 100 g *0.380 J^oC^-1g(80-T)

by principle of calorimetry

heat gained by cold body =heat lost by hot body

q_water =q_copper

2500g *4.183 JC^-1g*(T-25)=100 g *0.380 J^oC^-1g(80-T)

10457.5*(T-25)=38(80-T)

10457.5 T-261437.5 = 3040-38T

10457.5 T+38T = 3040+261437.5

10495.5T =264477.5

T = 264477.5/10495.5= 25.19⁰ C answer