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-Dominant- [34]
3 years ago
8

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

f heat could be transferred to sore muscles?
Physics
2 answers:
IgorC [24]3 years ago
8 0

Answer:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

Explanation:

For this case we know the mass of the water given :

m = 787 gr

And we know that the initial temperature for this water is T_i =75 C.

We want to cool this water to the human body temperature T_f = 37 C

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Q= m c_p \Delta T

Where c_p represent the specific heat for the water and this value from tables we know that c_p =1 \frac{cal}{gr C} for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

alexandr402 [8]3 years ago
4 0
<h2>Answer:</h2>

125.007 kJ

<h2>Explanation:</h2>

The quantity of heat (Q) transferable in a heating process is the product of the mass (m) of the object involved, the specific heat capacity of the object and the change in temperature (ΔT). i.e

Q = m x c x Δ T     ------------------------(i)

Where, from the question;

m = mass of water = 787g

c = specific heat capacity of water = 4.18 J/g°C  ---- (a known constant)

ΔT = 75°C - 37°C = 38°C

<em>Substitute these values into equation (i)</em>

Q = 787 x 4.18 x 38

Q = 125007.08 J

<em>Convert to kilojoules;</em>

Q = 125.007 kJ

Therefore, the quantity of heat (in kilojoules) that could be transferred to sore muscles is 125.007

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