Question

After transfer of 1.50 kJ of thermal energy to a 0.663-kg block of copper the temperature is 47.8 °C. The specific heat capacity of copper is 0.385 J g-1 °C-1.

Calculate the initial temperature of the copper.

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Answer 1

`Q=cm\Delta T\\ \Delta T=T_(f)-T_(i)\\\\ Q=cm(T_(f)-T_(i)) \ \ \ |:cm\\\\ T_(f)-T_(i)=(Q)/(cm)\\\\ T_(i)=T_(f)-(Q)/(mc)\\\\ T_(f)=47,8^(o)C\\ Q=1,5kJ=1500J\\ m=0,663kg=663g\\ c=0,385(J)/(g^(o)C)\\\\\\ T_(i)=47,8^(o)C-(1500J)/(663g*0,385(J)/(g^(o)C))=47,8^(o)C-5,88^(o)C=41,92^(o)C`

Answer 2

The initial temperature of the copper was 41.9 °C

Step-by-step explanation:

Data:

Heat transferred, Q = 1.5 kJ = 1500 J

Mass of copper, m = 0.663 kg = 663 g

Final temperature, T2 = 47.8 °C

Specific heat capacity of copper, cp = 0.385 J/(g °C)

Initial temperature, T1 = ? °C

The heat transferred is computed as follows:

Q = m*cp*(T2 - T1)

Isolating T1:

T1 = T2 - Q/(m*cp)

Replacing with data (units are omitted):

T1 = 47.8 - 1500/(663*0.385)

T1 = 41.9 °C