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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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Verified Answer :

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

Answer:

The initial temperature of the copper was 41.9 °C

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 = mcp(T2 - T1)

Isolating T1:

T1 = T2 - Q/(mcp)

Replacing with data (units are omitted):

T1 = 47.8 - 1500/(6630.385)

T1 = 41.9 °C

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