760 J of heat is used to increase the temperature of 400g of copper block from 30 ℃ to 35 ℃. What is the specific heat capacity of the copper block?
Q=cm(t2−t1)→c=Qm(t2−t1)=7600.4⋅(35−30)=380 (J/(kg⋅K))Q=cm(t_2-t_1)\to c=\frac{Q}{m(t_2-t_1)}=\frac{760}{0.4\cdot(35-30)}=380\ (J/(kg\cdot K))Q=cm(t2−t1)→c=m(t2−t1)Q=0.4⋅(35−30)760=380 (J/(kg⋅K)) . Answer
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