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

Answer:

$T_{final} =45.64C$=final temperature

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 ,

$q=m\times c \times \Delta T$

$\Delta T=\frac{q}{mc} =\frac{1962}{732 \times 4.18}$

$\Delta T= 0.64C$

$T_{final} -T_{initial} =0.64C$

$T_{final} =45.64C$=final temperature.