Question

A 30.0-g sample of water at 280. K is mixed with 50.0 g water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Answer 1

Answer:

311.25 K

Explanation:

According to principle of calorimetry,

Heat lost by hot body = Heat gained by cold body

m1c1Δt=m2c2Δt

c is the same. So, it gets cancelled.

50*(330-t)=30*(t-280)

16500-50t=30t-8400

16500+8400=30t+50t

24900=80t

80t=24900

t=24900/80

=311.25 K

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Answer 2

Given:

A 30.0-g sample of water at 280. K is mixed with 50.0 g water at 330. K.

To find:

Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

Explanation:

$q=m.c$Δ$T$

$q$ - heat lost or gained

$m$ - the mass of the sample

$c$ - the specific heat of the substance

Δ$T$  - the change in temperature, defined as final temperature minus initial temperature.

$-q lost=q gained$

$m_{hot}.c.$Δ$_{Thot}$$=m_{cold}.c.T_{cold}$

$-50.0_{g}.(T_{f}-330)K$$=30.0_{g}.(T_{f}-280)K$

$-50.0.T_{f}+16500$$=30.0.T_{f}-8400$

$80.0.T_{f}=24900$

$T_{f}=$$\frac{24900}{80.0}$$=311.25K$

$T_{f}=310 K$

Answer:

Therefore, The final temperature of the mixture heat loss is $T_{f}=310K$.