Question

Consider two object has mass m1=5.0kg m2=0.5kg and the specific heat c1=c2=4200j/kg oc respectively assume that first object has temperature 20oc and second object has temperature 90oc After they place these two object touch each other determine the mixture temperature of the substance

Answer 1

Answer:
`T=26.36^(\circ)C`

Step-by-step explanation:

Given

mass of the first object
`m_1=5\ kg`

mass of the second object
`m_2=0.5\ kg`

specific heat of the two is
`c_1=c_2=4200\ J/kg/^(\circ)C`

The temperature of the first
`T_1=20^(\circ)`

The temperature of the second
`T_2=90^(\circ)`

Heat flows from high temperature to low temperature

Suppose T is the common temperature

`m_1c_1(T-T_1)=m_2c_2(T_2-T)\\5* 4200* (T-20)=0.5* 4200* (90-T)\\5T-100=45-0.5T\\5.5T=145\\T=26.36^(\circ)C`