Question

4. When 732 grams of water was heated, it absorbed 1962 Joules of heat. The original temperature of the water was 45°C. What was the final temperature?

Answer 1

`T_(final) =45.64C`=final temperature

Step-by-step explanation:

In the question specific heat of water is not given but we should know the value of that and it 4.18Jg∘C

Specific heat means how much heat is required to increase the temperature of 1 gram of substance that substance by 1∘C .

Equation between heat lost or gain and the change in temperature.

q=m⋅c⋅ΔT , where

q - the amount of heat

m - the mass of the sample

c - specific heat of sample

ΔT - change in temperature

put all the given value into this ,

`q=m* c * \Delta T`

`\Delta T=(q)/(mc) =(1962)/(732 * 4.18)`

`\Delta T= 0.64C`

`T_(final) -T_(initial) =0.64C`

`T_(final) =45.64C`=final temperature.