Question

What is rise in temperature of 1 kg water if 1000 j heat is supplied?

Answer 1

$\Huge{\red{\underline{\textsf{Answer}}}}$

Answer 2

Answer:

We C=4.186C=4.186 Joule / gram

\therefore∴ \triangle T=\dfrac { Q }{ C\times m } =\dfrac { 1000 }{ 4.186\times 1000 }△T=

C×m

Q

=

4.186×1000

1000

={ \left( \dfrac { 1000 }{ 4186 } \right) }^{ 0 }C=(

4186

1000

)

0

C