Question

Question #85239

Answer 1

Pretty Huuugeee!

An edited version was added below which yielded an answer of `57.143 ^@K`.

Explanation:

The question provided a value of specific heat capacity with incorrect units. In the answer immediately below, Gaya3 made an assumption that a value of `C=3.5 "J""kg"^-1"K"^-1` was intended. The answer, using that value for C, was 57143K.

I, Steve J, am making a different interpretation of what was intended for C. I have added Answer 2 , below Gaya3's work (and using that format for the math), a solution using `C=3.5 "kJ""kg"^-1"K"^-1`. The difference being that the J is changed to kJ.

Answer 1
`q=mC∆T`

Here,

• `q= 2MJ =2times10^6J`
• `m=10kg`
• `C=3.5 "J""kg"^-1"K"^-1`
• `∆T=?`

So, `2times10^6J=10cancel(kg) times3.5 "J"cancel("kg"^-1)"K"^-1times∆T`

`=>∆T=(2times10^6cancel(J))/(10 times 3.5 cancel("J")"K"^-1)~~57143K`

Answer 2

`q=mC∆T`

Here,

• `q= 2MJ =2times10^6J`
• `m=10kg`
• `C=3.5 " kJ""kg"^-1"K"^-1 = 3.5 times10^3"J""kg"^-1"K"^-1`
• `∆T=?`

So, `2times10^6J=10cancel(kg) times3.5 times 10^3"J"cancel("kg"^-1)"K"^-1times∆T`

`=>∆T=(2times10^6cancel(J))/(10 times 3.5 times 10^3cancel("J")"K"^-1)~~57.143^@K`