a hot solid of mass 60 grams at 100 degrees celsius is placed in 10 gram of water at 18 degree celsius. the final study temperatureis recorded as 20degree celcius find the specificheat capacity of the solid

Since you want to assume no heat energy is lost from the system, make the two equal to each other in the equation q=mct.
M being mass
C being specific heat
T being change in temperature

(60g)*(c)*(80C) = (10g)*(4.18J/C*g)(2C)

80 is the change in temperature of the solid 100-80= 20

2 is the change in temperature of the water 20-18=2

4.18 is the specific heat of water and you should memorize it because it’s expected of you

So now solve for c:
(4,800g*C)c = 83.6J
c = 0.0174 J/g*C