Question

`When 800 J of heat is given to 200 g sample of a metal its temperature increased by 40 °C. Il specific heat of metal is m x 50 J kg 1.c 1 then find the value of m. When 800 Joule of heat is given to 200 g sample of a metal its temperature increased by 40°C. If specific hoat of metal is m x 50 J kg then find the value of m. When 800 J of heat is given to 200 g sample of a metal its temperature increased by 40 °C. If specific heat of metal is m x 50 J kg-1.c- then find the value of m. `

I was given this question in fiitzee exam, I'm in class 7, please explain in simple way

please answer fast i will mark as brainliest

Answer 1

Answer: value of m is 1.25

Explanation: in first case ,Q=ms(t2-t1) ,, here temperature difference is 40

so after putting value of the parameter we get the value of m is 1.25

Answer 2

Answer: value of m is 1.25

Explanation: in first case,Q=ms(t2-t1),, here temperature difference is 40

so after putting value of the parameter we get the value of m is 1.25