The heat gained is given by;
"\u0394H=m *s*(t2\u2212t1)"
Where ΔH=305J
Mass of metal, m=68.6g
Specific heat of metal, s=0.128Jg∘C
t1→initial temperature=20∘C
t2→final temperature=?
Therefore,"\u0394H=m*s*(t2\u2212t1)"
305J= 68.6g×0.128Jg∘C×(t2−20)
305=8.7808t2-175.616
t2=480.616/8.7808
t2=54.734876093
Hence t2=54.73∘C
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